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J. D. Ryder / Charles M. Thomson 




Prentice 
Hall 



Electronic Circuits 
and Systems 



Electronic Circuits 
and Systems 



J. D. Ryder 



Formerly Professor of Electrical Engineering 

Michigan State University 

East Lansing. Michigan 



Charles M. Thomson 

Formerly Senior Dean of Instruction 

Wentworth Institute and 

Wentworth College of Technology 

Boston, Massachusetts 



PRENTICE-HALL, INC., Englewood Cliffs, New Jersey 






Library of Congress Cataloging In Publication Data 

Ryder, John Douglas (date) 

Electronic circuits and systems. 

Includes index. 

1. Electronics. 2. Semiconductors. 3. Electronic 

circuits. I. Thomson. Charles M., joint author. 

II. Title. 

TK781S.R89 

ISBN 0-13-250407-3 




1 3 NOV 1987 



© 1976 by PRENTICE-HAL 
Englewood Cliffs, New Jersey 



All rights reserved. No part o 
may be reproduced in any for 
without permission in writing 



10 987654321 



Printed in the United States of America 



PRENTICE-HALL INTERNATIONAL, INC., London 
PRENTICE-HALL OF AUSTRALIA, PTY. LTD., Sydney 
PRENTICE-HALL OF CANADA, LTD., Toronto 
PRENTICE-HALL OF INDIA PRIVATE LIMITED, New Delhi 
PRENTICE-HALL OF JAPAN, INC., Tokyo 
PRENTICE-HALL OF SOUTHEAST ASIA (PTE.) LTD., Singapore 



Contents 



: 



Preface 



xiii 



CHAPTER 



The Semiconductor Diode 



1.1 Eleclrons; Photons I 

1.2 Atoms 2 

1.3 Crystals 3 

1.4 Mclals, Insulators, and Semiconductors 4 

1.5 Conduction in Silicon and Germanium 5 

1.6 The Forbidden Energy 7 

1.7 n and p Semiconductors 8 

1.8 Purification of a Semiconductor 10 

1.9 Manufacture of a pn Junction 1 1 

1.10 The Junction Diode 1 1 

1.11 The Diode Voliage-Currenl Equation 13 

1.12 Zener Diodes 15 

1.13 Light-Emitting Diodes. 16 

1.14 Photodiodcs 16 

1.15 Capacitance Effects in the pn Diode 17 

1. 16 General Comments 17 



vi Contents 

CHAPTER d£ The Diode as Rectifier and Switch 21 

2.1 The Ideal Diode Model 21 

2.2 The Half-Wave Rectifier Circuit 22 

2.3 The Full-Wave Rectifier Circuit 24 

2.4 The Bridge Rectifier Circuit 25 

2.5 Measurement of the Ripple in the Rectifier Circuit 26 

2.6 The Capacitor Filter 27 

2.7 The n Filter 31 

2.8 The n-R Filter 34 

2.9 The Voltage-Doubling Rectifier Circuit 35 

2.10 Rectifying AC Voltmeters 37 

2.11 Diode Wave Clippers 38 

2.12 Diode Clampers 41 

2.13 Review 42 



CHAPTER O Models for Circuits 48 

3.1 The Black Box Concept 48 

3.2 Active One-Port Models: the Voltage-Source Circuit 50 

3.3 Active One-Port Models: the Current-Source Circuit 51 

3.4 Maximum Power Output 53 

3.5 The Two-Port Network 54 

3.6 The //-Parameter Equivalent Circuit 56 

3.7 Power in Decibels 58 

3.8 Comments 60 



CHAPTER *T Junction Transistors as Amplifiers 64 

4.1 Transistor Voltage and Current Designations 64 

4.2 The Junction Transistor 65 

4.3 The Reverse Saturation Current in Transistors 67 

4.4 The Volt-Ampere Curves of a Transistor 67 

4.5 The Current Amplification Factors 69 

4.6 Relations Between the Amplification Factors 71 

4.7 The Load Line and Q Point 72 

4.8 The Basic Transistor Amplifiers 74 

4.9 Simplification of the Equivalent C-E Circuit 75 



Contents 



vii 



4.10 The Transconductancc g m 77 

4.11 The Common-Emitter Amplifier 78 

4.12 Performance or a C-E Amplifier 80 

4.13 Relation Between A, and A,. 81 

4.14 Conversion of the /; Parameters 81 

4.15 The Common-Base Transistor Amplifier 82 

4.16 Performance of the C-B Amplifier 84 

4.17 The Common-Collector Amplifier 85 

4.18 Performance of the C-C Amplifier 87 

4.19 Comparison of Amplifier Performance 88 

4.20 Transistor Manufacturing Techniques 89 

4.21 Comments 92 



CHAPTER 



DC Bias for the Transistor 



96 



5.1 Choice of the Quiescent Point 96 

5.2 Variation of the Q Point 99 

5.3 Fixed Transistor Bias 100 

5.4 The Four- Resistor Bias Circuit 102 

5.5 Design of a Fixed-Bias Circuit 105 

5.6 Design of a Bias-Stabilized C-E Amplifier 107 

5.7 Voltage Feedback Bias 1 10 

5.8 Design of a Voltage-Feedback Bias Circuit 1 12 

5.9 Bias for the Emitter Follower 112 

5.10 Design of the Emitter Follower Circuit 1 14 

5.11 Comments 115 



CHAPTER O The Field-Effect Transistor 

6.1 The Junction Field-Effect Transistor 120 

6.2 The MOS Field-Effect Transistor 123 

6.3 Symbols for the FET 125 

6.4 The Load Line for the FET 125 

6.5 Obtaining Bias for the FET 127 

6.6 The Four-Resistor Bias Network 129 

6.7 Design Examples 130 

6.8 The FET as an Amplifier 132 

6.9 Circuit Characteristics of the FET 132 



120 



viii 



CHAPTER 



Contents 



6.10 The Equivalent Circuit or the FET 135 

6.11 The Common-Source Amplifier 136 

6.12 The Common-Drain Circuit 137 

6.13 Design Example 140 

6.14 Using the FET as a Variable Resistance 142 

6.15 Summary 142 



CHAPTER 



The Vacuum Tube 



145 



7.1 

7.2 

7.3 

7.4 

7.5 

7.6 

7.7 

7.8 

7.9 

7.10 

7.11 



145 



Circuit Notation for the Vacuum Tube 

The Triodc 146 

Triode Characteristics 148 

The Pentode 150 

The Equivalent Circuit 152 

The Triode Common-Cathode Amplifier 153 

The Pentode Common-Cathode Amplifier 154 

The Cathode Follower 155 

The Grounded Grid Amplifier 157 

The Cathode-Ray Tube 159 

Comments 161 



Contents 



CHAPTER 



ix 



O Frequency Response of RC Amplifiers 



164 



8.1 
8.2 
8.3 
8.4 
8.5 
8.6 

8.7 

8.8 

8.9 

8.10 

8.11 

8.12 

8.13 

8.14 



Cascaded Amplifiers 164 

The Amplifier Passband 165 

The Frequency Plot 167 

Low-Frequency Response 168 

The Low-Frequency Limit 172 

Low-Frequency Response for the FET and Vacuum-Tube 

Amplifiers 175 

The Unbypassed Emitter Resistor 177 

High-Frequency Equivalent Circuits; the Miller Effect 180 

High- Frequency Response 182 

The Frequency Limit of the Transistor 185 

The Common-Base Connection at High Frequencies 188 

Bandwidth of Cascaded Amplifiers 189 

Frequency Emphasis and De-Emphasis 191 

Review 194 



' Negative Feedback in Amplifiers 199 

9.1 The Black Box with Feedback 199 

9.2 Stabilization of Gain by Negative Feedback 202 

9.3 Bandwidth Improvement with Negative Feedback 204 

9.4 Reduction of Nonlinear Distortion 206 

9.5 Control of Amplifier Output and Input Resistances 207 

9.6 A Current Series-Feedback Circuit 209 

9.7 Voltage Shunt-Feedback Circuit 212 

9.8 Voltage Feedback with the FET 213 

9.9 The Emitter Follower as a Feedback Amplifier 215 

9.10 Multiple-Stage Feedback 216 

9.11 Amplifier Gain Stability with Feedback 217 

9.12 Gain and Phase Margin 220 

9.13 Comments 221 

CHAPTER 10 Integrated Amplifiers • 225 

10.1 The Integrated Amplifier 225 

10.2 The Differential Amplifier 228 

10.3 Rejection of Common-Mode Signals 232 

10.4 A Constant-Current Circuit for R R 236 

10.5 Voltage References 237 

10.6 The DC Level Shifter 237 

10.7 The Operational Amplifier 238 

10.8 The Unity-Gain Isolator 242 

10.9 The Summing Operation 242 

10.10 The Integration Operation 243 

10.11 The Comparator 245 

10.12 A Millivoltmctcr 245 

10.13 Frequency Compensation 246 

10.14 The Slew Rate 247 

10.15 Offset Voltage and Current 249 

10.16 Definition of Terms 249 

10.17 The Darlington Compound Transistor 250 

10.18 The Cascode Amplifier 252 

10.19 Silicon Integrated Circuits 253 

10.20 Passive Elements 255 

10.21 Comments 256 



Contents 



CHAPTER 



A. A. Tuned and Video Amplifiers 

11. 1 Band-Pass Amplifiers 261 

11.2 The Parallel-Resonant Circuit 263 

11.3 Bandwidth of the Resonant Circuit 265 

11.4 The Single-Tuned Transistor Amplifier 268 

11.5 The Double-Tuned Transformer 269 

11.6 The Pulse Waveform 271 

11.7 The Shunt-Peaked Video Amplifier 273 

11.8 The Series-Peaked Video Amplifier 277 

11.9 Review 278 



261 



CHAPTER 



J. ^ Power Amplifiers 

12.1 Defined Operating Conditions 283 

12.2 The Ideal Transformer 285 

12.3 Power Relations in the Class A Amplifier 286 

12.4 Voltage Limitations 290 

12.5 Effect of the Thermal Environment 290 

12.6 Determination of Output Distortion 293 

12.7 The Push-Pull Circuit and Class B Operation 297 

12.8 Performance of a Class B Push-Pull Amplifier 300 

12.9 Output Circuits Without Transformers 302 

12.10 Phase Inverters for Push-Pull Input 304 

12.11 Complementary Symmetry Circuits 305 

12.12 The Class B Linear Radio-Frequency Amplifier 307 

12.13 Summary 308 



283 



CHAPTER 



J. yj Oscillator Principles 

13.1 Oscillator Feedback Principles 313 

13.2 The Hartley and Colpitts Oscillators 316 

13.3 Practical Transistor Oscillators 316 

13.4 Crystal Control of Frequency 319 

13.5 Resistance-Capacitance Feedback Oscillator 322 

13.6 Comments 323 



313 



Contents 



xi 



CHAPTER 



14 



Modulation and Detection 



326 



14.1 Fundamentals of Modulation 326 

14.2 The Frequency Spectrum in AM 327 

14.3 The Power Spectrum in AM 330 

14.4 The Diode Modulator for AM 331 

14.5 High Power-Level AM Modulation 333 

14.6 Linear Detection for AM Signals 334 

14.7 Automatic Volume Control (AVC) 336 

14.8 The Single-Sideband System of Modulation 337 

14.9 Frequency Translation; the Product Detector 339 

14.10 The Frequency Spectrum of FM Signals 341 

14.11 Bandwidth of FM Signals 342 

14.12 Generation of FM Signals 344 

14.13 The Limitcr and Discriminator for FM Detection 344 

14.14 The Ratio Detector for FM 347 

14.15 Automatic Frequency Control (AFC) 348 

14.16 Comments 349 



CHAPTER 



15 



Radio Systems 



3S3 



15.1 Noise 353 

15.2 Information in Signals 354 

15.3 Information Capacity of a Channel 355 

15.4 An AM Transmitter 357 

15.5 The Superheterodyne Receiver 358 

15.6 The SSB Transmitter 362 

15.7 An SSB Transceiver 363 

15.8 AM Versus FM 366 

15.9 FM Systems 367 

15.10 A Radar System 368 

15.11 Frequency Classification for Radio Signals 371 



CHAPTER 



16 



Digital Circuits 



374 



16.1 Binary Numbers 374 

16.2 Binary Arithmetic 377 

16.3 Binary Codes 378 

16.4 Other Number Systems 379 






xii 




CHAPTER 



16.5 

16.6 

16.7 

16.8 

16.9 

16.10 

16.11 

16.12 

16.13 

16.14 

16.15 

16.16 

16.17 

16.18 

16.19 

16.20 

16.21 

16.22 

16.23 

16.24 

16.25 

16.26 

16.27 

16.28 

16.29 

16.30 

16.31 

17 

17.1 

17.2 
17.3 
17.4 
17.5 
17.6 
17.7 



Contents 

Scries and Parallel Processing of Bits 381 

Logic Operations in Addition 382 

Logic Switches 383 

Logic Voltage Levels 386 

Diode Logic (DL) Gages 386 

The not or Inversion Operation 388 

Integrated Logic Circuits 389 

Diode-Transistor Logic (DTL) 390 

Resistance-Transistor Logic (RTL) 392 

Transistor-Transistor Logic (TTL) 393 

Emitter-Coupled Logic (ECL) Circuits 394 

CMOS Logic Circuits 395 

nand and nor Circuits with CMOS Logic 396 

CMOS Manufacture 398 

The Adder Circuit as a Logic System 399 

Multivibrators 400 

The Bistable Multivibrator or Flip-Flop 401 

The RS Flip-Flop 402 

The T or Toggling Flip-Flop 403 

The JK Flip-Flop 405 

The One-Shot or Monostablc Multivibrator 407 

The Astable Multivibrator 408 

Synchronization of the Free-Running Multivibrator 410 

The Schmitt Trigger Circuit 41 1 

The Shift Register 412 

Decoding Matrices 413 

Decimal Counting 414 



Power Control 

The Silicon-Control Rectifier (SCR) 422 

SCR Characteristics 423 

The Unijunction Transistor (UJT) 425 

Triggering the SCR 427 

The Controlled Polyphase Rectifier 430 

Shunt- Wound DC Motor Control 431 

Comments 432 

Index 



422 



Preface 



435 



This book has been written to provide the electronic technician with a 
broad and basic understanding of the diode and transistor and other devices, 
the circuits, and the systems in which they are employed in the electronics 
field. The technician is often most interested in practical electronic applica- 
tions in the real world. To foster this interest we have included selected appli- 
cations with the solid-state and transistor theory. 

The concept of electrical equivalent circuits or black boxes is developed 
early in the text and consistently applied. To demonstrate the similarity of the 
active electronic devices, only one circuit model is used to describe all three 
amplifying devices, the bipolar transistor, the field-effect transistor, and the 
vacuum lube. The circuit model employed is the g m -controlled constant-cur- 
rent source. It is simplified in form for easy understanding of the circuit opera- 
tion, but is sufficiently accurate for basic circuit design procedures. 

The material presented, and the terminology employed, is intended to 
acquaint the student with practical usage in the field. The mathematics 
involved is restricted to algebra and simple trigonometry. Derivations arc 
rather thoroughly developed. In this manner the student can discover how 
algebra is employed in reasoning and leads to results which have electrical 
meaning. Many worked examples provide the student with a knowledge of 
the circuit conditions encountered under a variety of operating conditions. It 
is expected that the student will have completed or is currently taking a course 
in ac circuit fundamentals. 



xiii 



xiv 



Preface 



Fundamental material on the solid-state conduction process is introduced 
in Chapter I . This provides curious students with answers to the why and how 
of diodes and transistors. 

Diodes result from the ability to transform silicon and germanium into n 
and p forms, and diode theory and application are introduced in Chapter 2. 
Rectifier circuits arc presented, along with the function of filter circuits. This 
material is introduced here since it demonstrates an important application of 
the diode, as well as furnishing source material for early laboratory experi- 
ments. 

The idea of a black box as an equivalent model of an electrical circuit or 
device is introduced in Chapter 3. One-port and two-port forms are discussed, 
and the equivalent circuit /(-parameters are brought forward. 

In Chapter 4, the operation of the transistor is developed, and the equiva- 
lence between the transistor and the /j-parameter two-port circuit is then 
shown. Thereafter the common-emitter h parameters and the g m model are 
used for amplifier analysis throughout the book. The low-frequency perfor- 
mance of the three basic amplifier forms is developed and compared. Tech- 
niques used in the manufacturing of several modern forms of the transistor 
arc discussed. 

The necessity for choice of an operating point in the mid-region of the 
transistor output curves to obtain linear and undislorted response is shown in 
Chapter 5. Bias circuits which establish and maintain this operating point are 
found. Several circuits are treated as design examples. 

The important field-effect transistor is covered in Chapter 6. The g m cur- 
rent-source model is used again. Bias circuits and complete amplifier designs 
are developed. 

Building on the previous circuit theory, and again employing the g m 
model, the vacuum tube is introduced in Chapter 7. Triode and pentode 
circuits are briefly discussed. The cathode-ray lube is also introduced here. 

The response of amplifiers over broad frequency ranges is considered in 
Chapter 8 and the problems which arise in multistage amplifiers are also 
introduced. The efTect of the internal parasitic capacitances, always present 
in all devices, on the response of the amplifying devices at high frequencies 

is discussed. 

Feedback is of the utmost importance in most electronic applications, and 
sufficient background has been developed to permit a discussion of feedback 
in Chapter 9. The performance improvements obtained with negative feed- 
back are shown, and we also introduce some of the problems of amplifier 
instability. Feedback operation in multistage amplifiers is given special atten- 
tion. 

The universally used integrated or monolithic differential amplifiers are 
presented in Chapter 10. A discussion of the basic circuits within the usual 



Preface 



amplifier, and the methods of manufacture, follows. The operational ampli- 
fier, and a few of its many applications are included. 

Chapter 1 1 provides coverage of the frequency response of transistors in 
tuned radio-frequency amplifiers. Also included are the rectangular pulse 
wave form and its use in video amplifiers or in data circuits. 

Power amplifiers and their associated heat-removal problems are consid- 
ered in Chapter 12. General push-pull circuits as well as modern transfor- 
merless circuits are covered. The class B linear radio-frequency amplifier is 
also discussed because of its popularity in increasing power at radio frequen- 
cies. 

Feedback is used to create the several forms of tuned oscillator circuits, 
crystal oscillators, and the laboratory form of RC oscillator presented in 
Chapter 13. 

Having established a foundation of various building-block electronic cir- 
cuits, we proceed to a study of modulation and detection of amplitude-modu- 
lated and frequency-modulated signals in Chapter 14. Frequency spectra and 
bandwidth requirements for the various signals are shown, as well as methods 
of generation of the modulated signals. 

Fn Chapter 15 the concepts of information content of a signal and the 
information capacity of a channel are introduced. The previously-developed 
basic circuits are combined in several complete radio systems, as examples. 

Chapter 16 departs from the realm of the continuous or analog signal and 
introduces the binary code, the binary pulse signal and logic circuits for data 
manipulation and computation. Modern ECL, TTL, and CMOS forms of 
logic circuits are included. Logic circuit applications in adding, flip-flop and 
counting circuits are used as examples. 

The text concludes with Chapter 17 which details the characteristics of 
power-switching devices and their applications in several areas of power 
control. 

Questions to encourage student review and extensive lists of problems are 
supplied at the end of each chapter. 

John D. Ryder 
Charles M. Thomson 






Electronic Circuits 
and Systems 



1 

The Semiconductor 

Diode 



In recent years we have learned a great deal about electrical conduction in 
solids and have been able to make semiconductor materials having electrical 
properties that normally do not occur in nature. From this research into the 
electrical properties of solids, the pn junction diode has been developed. It 
is widely applied as a rectifier and as a device with a nonlinear voltage- 
current characteristic. Principles found in the development of the diode are 
also employed in the transistor. 



1.1 Electrons; Photons 

Electronic science had its beginning in 1883 when Edison observed that a 
current would pass between a metal plate and the heated filament in one of 
his lamps if the plate was made positive in relation to the filament. This 
"Edison effect" was due to the flow of negative particles, later named elec- 
trons. The electron is now accepted as the fundamental unit of electric charge, 
negative in nature and designated as —e. It is so small that 6.25 x 10 2s 
electrons must pass per second to represent a current of one ampere. 

The physical form of an electron is unknown but we shall assume that it 
is a spherical particle with a mass measured as 9.11 X 10" 3 ' kilogram. 

The photon is a bundle of radiant energy appearing as light, heat, X rays, 
and other electromagnetic radiation. The size of the energy package is in- 
versely related to the wavelength of the radiation and is measured in joules. 



The Semiconductor Diode 



1.2 Atoms 



The chemical elements are built of atoms. The Bohr theory of atom.c struc- 
ture proposes a central nucleus of positive electrical charge and mass, around 
which electrons move in definite orbits. The positive charge of the nucleus 
is due to protons, each having a positive charge equal to that of the electron. 
The proton has a mass approximately equal to that of the hydrogen atom. 
Additional mass is added to the nucleus of heavier elements by the presence 
or uncharged particles known as neutrons. 

Each element has a number of protons in its nucleus corresponding to 
its atomic number, which can be found in the periodic table. A normal atom 
is electrically neutral since the number of protons in the nucleus .s equal to 
the number of electrons in the surrounding orbits. 

There is a fixed amount or level of rotative energy associated with each 
orbif electrons in the innermost orbits close to the nucleus have the least 
energy and those at greater distance from the nucleus have greater energy. 
Orbits with common energy levels are called shells or rings. Each shell or 
ring has positions for a definite number of electrons. When these posit.ons 
are occupied by electrons, the shell is said to be filled. 

The inner shells are normally filled and their electrons are shielded from 
external forces by the charges of the outer shell electrons. The electrons in 
the outermost shell are called the wlenee electrons since it is these outer 
electrons that contact the neighboring atoms and give the element its expected 
properties in forming chemical compounds. 

Hydrogen has only one proton and one electron, and possession of a 
single electron gives hydrogen a valence of one and places it in Group I of 
the periodic table. Copper, another Group I element with one valence elec- 
tron in its outer shell, has its 29 electrons grouped as 



Shell K L 

No. of electrons 2 8 



M 
18 



N 

1 



The elements found in Group I have many unfilled positions in the outer 
shell and easily join with other elements having greater numbers of valence 
electrons. Thus Group I elements are considered chemically active 

Elements of Group VIII such as helium, neon, and argon have filled outer 
shells with eight electrons. There are no unfilled energy levels available to 
valence electrons of other atoms and Group VIII elements are chemically 

inactive. . ... 

The Group IV elements include our important semiconductors, silicon 
and germanium, with the following electron arrangement in the shells: 



Crystals 



K L M N 

Carbon 2 4 

Silicon 2 8 4 

Germanium 2 8 18 4 



The four electrons in the valence shell provide desirable semiconductor 
properties. Even carbon becomes a semiconductor when in diamond form 
and heated sufficiently. 

An atom bombarded by a high velocity electron or other particle carrying 
sufficient energy may have an electron knocked out of the valence shell. 
This leaves the atom with a net positive charge of Ye, and the positively 
charged atom is known as a positive ion. The energy supplied by the bombard- 
ing particle is known as the ionizing energy. 

1.3 Crystals 

In a liquid there is a random location of the atoms but as an element freezes 
from the liquid state, the distance between atoms decreases and the binding 
forces increase. This gives the material its strength as a solid, in many solids 
the atoms assume regular, lattice-like arrangements much as spheres pack in 
a box in layered order. The resultant structural arrangement of atoms is 
called a crystal. Crystals repeat the unit structure, shown in Fig. l.l, as well 
as other forms. The form for a given clement is determined by the nature of 
the forces that bind the atoms in the solid. 



/ 

i— 



A" 



y 



&>C 



n 



y— +-> 



/ 



i 









*- 



(a) 



trfr— X* /. 

! *H / k ! 

■0sr 

(b) 






*L 



w 



(c) 



Figure /./ Unit crystal cells: (a) simple cubic lattice: (b) facc-ccnlercd cubic 
lattice; (c) body-centered cubic lattice. 



Mechanical working of a material, such as the stretching of a wire, tends 
to create a structure of many small crystals. The random boundaries of these 
individual crystals cause erratic electrical conduction properties. Semi- 
conductor materials are carefully grown in single crystal form, however, and 
this ensures that crystal boundaries do not disturb the desired conduction 
properties. 



The Semiconductor Diode 



1.4 Metals, Insulators, and 
Semiconductors 

Our good electrical conductors are metals, such as silver, copper, or gold. 
In metals the interatomic spacing is so close that the orbits of the outer- 
most electrons of neighboring atoms overlap and a valence electron will be 
in the attractive fields of several nuclei. The fields essentially cancel, leaving 
these outer electrons very loosely held. They are freed from their nuclei by 
the random thermal vibrations of the atoms at room temperature. Copper 
and silver have one valence electron and thus we have one electron per atom 
available to move as a free charge. These metals have about 10 23 atoms per 
cm 3 and so we have 10 23 electrons per cm 3 as free charges available to move 
in the conduction process.'" 

When we apply a voltage to a wire, individual charges move in a random 
manner toward the positive terminal at low velocity. With the large number 
of free charges we find that the electrical resistivity of metals is low (see 
Table 1.1). 



TABLE 1.1 Electrical Properties of Materials 



Material 



Properly 



Resistivity 
(fl-m) 



Silver 


Conductor 


1.6 x 10-8 


Germanium 


Semiconductor 


0.6 


Silicon 


Semiconductor 


1500 


Polystyrene 


Insulator 


10'» 



There are other materials that bind their valence electrons so tightly that 
there are few charges able to move at the usual ambient temperatures. With 
few free charges these materials cannot carry an appreciable current and they 
are classed as insulators (quartz, porcelain, and polystyrene are examples). 

'We shall use the scientific method or notation, stating quantities as small numbers 
limes powers of 10, to avoid writing many zeros to the right or left of the decimal point. 
We shall also designate major magnitudes by the following prefixes, used with the unit 
name, as in w;//metcr ■ 0.01 meter: 



Multiple 


Prefix 


Symbol 


Multiple 


Prefix 


Symbol 


10'* 


tcra 


T 


10-2 


ccnti 


c 


10' 


giga 


G 


10- 3 


milli 


m 


10« 


mega 


M 


io-« 


micro 


p. (lowercase Greek mu) 


10' 


kilo 


k 


io-» 


nano 


n 








10- > 2 


pico 


P 



Conduction in Silicon and Germanium 5 

In pure form the semiconductors have conduction properties intermediate 
between the good conductors and the insulators. A pure semiconductor is 
an insulator at temperatures near absolute zero (— 460°F or — 273°C) but 
the resistance falls as the temperature rises. Such materials, whose resis- 
tance decreases with rising temperature, are said to have negative tempera- 
ture coefficients of resistance. 

Semiconductors of major importance are silicon and germanium; their 
properties are compared to a good conductor and to an insulator in Table 
1 . 1 . A number of metallic sulfides and oxides and compounds such as gallium 
arsenide and gallium phosphide are also useful in semiconductor applications. 



1.5 Conduction in Silicon 
and Germanium 



Our most important semiconductors, silicon and germanium, are from Group 
IV and they have four valence electrons. They form crystals having covalent 
bonds between atoms in which the valence electrons are shared in pairs with 
four adjacent atoms. This is diagrammed in Fig. 1.2. The electrons in the 
covalent bonds arc spinning on their axes but in opposite directions. The 
spin creates a magnetic field and the interlocking of these magnetic fields 

Electron 

, Covalent , [ 
/ Bond 




Nucleus 



■::::~M y ::: 



\ i 






i 



i ; 

i i 












i 
i i 



\ i \ i \ i 
n 



Figure 1.2 Schematic drawing of covalent bonds in germanium at absolute zero 
temperature. 



The Semiconductor Diode 
6 

provides the binding force of the covalcnt bond that holds the crystal 

t0g Absolute zero temperature (-460°F or -273°C) all the valence elec- 
trons of the semiconductor are tightly held in the covalcnt bonds w.th neigh- 
boring atoms. There are no charges free to move if a voltage is applied. 
Electrical conduction is not possible and the materials behave as msulators. 
As the temperature rises to the ambient range (SOT or 27°C), thermal energy 
is absorbed by the atoms and the electrons of the solid; this energy appears 
as random vibration or agitation of these particles about thc.r lattice loca- 
tions Some electrons acquire sufficient energy to break the covalent bond 
and the electrons become free and mobile. Energy for breakage of a covalent 
bond may also be supplied by a high voltage across the matenal or by rad.a- 
tion with photons of appropriate wavelength. 

When a bond is broken and an electron freed, an electron vacancy is left 
in the covalent bond. The vacant electron site is called a hole, indicated in 
Fig 1 3 The hole represents the absence of a negative charge and is attrac- 
tive to electrons; therefore the hole appears to have a positive charge of 
-\-e value. 






i 







* ♦ 
l I Nucleus 



I i Conduction 

♦ f * Klcctron f f* CoVJ 



*c P :» 



Covalent Bond 



' i K 

k . Hole * • 

1 ' 



Figure 1.3 Formation of a hole. 

An atom that has lost an electron is a positive ion fixed in the crystal 
lattice The total material remains electrically neutral, however, s.ncc the 
number of positive ions is still equal to the number of free electrons. 

The hole is also considered to be mobile. Examine Fig. 1.4, where a 
broken covalent bond has left a hole at A. Another electron may move from 
a bond at B to the hole at A, under the force of a small voltage applied across 
the material as shown. When the electron fills the hole at A there : is now a 
hole at B; then another electron may move from a bond at C to the hole at 
B and the hole has moved to C. In a similar manner an electron may move 
from a bond at D to the hole at C and so the hole has moved from A to D 
in stepwise fashion. 






The Forbidden Energy 



$$ <*><*> 



Original 
Hole 

/ \ Jump2 ^L \ ^«b1«25 



--€>- 



<•><$> V <•> 



•<\<°" 






'/>• 









\ 



D, 



JUmp3 <•)<•) <•)<*) 



Final Hole Location 



Figure 1.4 Conduction by a hole. 



The hole and the electron arc both mobile charges that can take part in 
electrical conduction in these materials. The positive hole progresses toward 
the negative terminal of an applied voltage, however, while the electron 
moves toward the positive terminal. In this text we define conventional 
electric current as a flow of positive charges so that the movement of holes 
corresponds to a conventional current. 

Since holes and free electrons are simultaneously produced when a 
covalent bond is broken, we speak of the process as the generation of elec- 
tron-hole pairs. This is a natural and inherent process in semiconductors 
and conduction with charges produced by pair generation is called intrinsic 
conduction. 

The number of pairs produced and the intrinsic conductivity are both 
low at room temperature since intrinsic conduction by holes and electrons 
is dependent on the thermal agitation of the atoms to provide the energy to 
break the valence bonds. The process is so sensitive to the temperature that 
intrinsic current levels may be expected to double for every increase of IO°C 
or 18°F, limiting the use of silicon devices to 390°F or 200°C, and germa- 
nium devices to 200°F or 90°C. 

Intrinsic conduction is not desired in transistor operation and the effect 
is suppressed by operation of those devices below these maximum tem- 
perature limits. 

1.6 The Forbidden Energy 

To become free and mobile, a valence electron in a semiconductor must 
acquire sufficient vibrational energy to break the covalent bond. The break- 



The Semiconductor Diode 



n and p Semiconductors 






age energy of a bond in a given material is a fixed quantity, called E .' 2 ' We 
cannot have a partially broken bond and so energies less than E are unac- 
ceptable or forbidden to the valence electrons. As a result, E c is called the 
forbidden energy or sometimes the gap energy in semiconductors. 

Measured in electron volts, the forbidden energies of the major semi- 
conductors are 



Silicon 
Germanium 



l.lSeV 
0.75 cV 



Since the forbidden energy of silicon is greater than that of germanium, more 
energy is needed to break a covalent bond in silicon and there will be fewer 
intrinsic charges. Thus silicon has lower intrinsic conductivity at ambient 
temperatures. 

1.7 n and p Semiconductors 

Germanium is obtained as a by-product in zinc refining and silicon is derived 
from the manufacture of silicones. Both materials arc highly purified to 
remove the random impurities of manufacture. 

To obtain the desired conduction properties, the pure silicon or german- 
ium is "doped" with minute amounts of selected elements as controlled 
impurities. Desirable doping elements include boron, aluminum, gallium, 
and indium; these have three valence electrons and are said to be trivalent 
impurities. Other doping elements are chosen from arsenic, antimony, and 
phosphorus; these have five valence electrons and are referred to as penta- 
valent impurities. 

The impurities are added at rates of one atom of doping element to 10 ! 
to 10 8 atoms of semiconductor. The small density of doping atoms causes 
them to be separated by thousands of semiconductor atoms in every direc- 
tion and the form of the crystal is not altered by the presence of the impurity 
atoms. An impurity atom can only find a place in the crystal by substituting 
for one of the germanium or silicon atoms; it cannot alter the crystal form. 

When a pcntavalent atom, such as arsenic, substitutes for a silicon atom, 
there are places in the covalent bonds for only four of its five valence elec- 
trons and one electron is left over. This is indicated in Fig. 1.5(a). The bind- 
ing force of this electron to its nucleus is slight and the thermal agitation at 
room temperature is sufficient to break this electron free. 

iWc measure Eo and other energies associated with atomic particles in electron rolls 
(cV). An electron receives I cV of energy when accelerated through 1 V; it receives 500 cV 
of energy when accelerated through a potential of 5C0 V. This numerical equivalence of 
potential and energy gained makes the electron volt a convenient measure of energy. 

The electron volt represents an energy of 1.60 x 10" >» joule. 



Excess 
Electron 








» ; 



/ \ 









Hole 



• * 



\ i 



\ I 






Figure I.S Semiconductor with (a) a pcntavalent arsenic impurity; (b) a trivalent 
indium impurity. 

By addition of a pentavalent impurity we have created a material having 
only electrons as free and mobile charges; such a semiconductor is said to 
be n material. The pentavalent atoms have donated an electron for conduc- 
tion and are called donor atoms. Since an electron has been removed from the 






10 



The Semiconductor Diode 



donor atoms, they remain as positive ions fixed in their crystal positions. 

When a trivalcnt atom substitutes for a semiconductor atom in a crystal, 
its three valence electrons enter the covalent bonds with electrons from three 
neighboring silicon or germanium atoms but an electron vacancy or hole is 
left in the bond to a fourth neighbor atom, as illustrated in Fig. 1.5(b). At 
room temperature the built-in hole can be filled by an electron from a nearby 
bond and this process creates a mobile hole in the adjacent area. 

The use of a trivalent impurity element has created a material with free 
holes so that conduction will occur by hole transfer and the impurity semi- 
conductor is said to be p material. The trivalent impurity atoms accept 
electrons to fill their bond vacancies and are called acceptor atoms. Since these 
atoms have acquired an extra electron, they remain fixed in the crystal as 
negative ions. 

In an n semiconductor the conduction will be predominantly by electrons 
and the electrons will be called the majority carriers. There will be a few holes 
present in an n material due to electron-hole generation at usual ambient 
temperatures. The holes present in an n material are known as the minority 

carriers. 

The conduction in a p material will be primarily due to holes and they 
are known as the majority carriers. There will be a few electrons present in 
a p material due to thermal-pair generation: these electrons in a p material 
are known as the minority carriers. 



1.8 Purification of a Semiconductor 



The purification of germanium or silicon to semiconductor standards is 
usually done by zone refining in an inert gas atmosphere. If the temperature 
of a short section of impure rod is raised to the molten state and the molten 
zone moved slowly, the impurities tend to remain in the molten zone and 
travel with it to the end of the rod. By multiple repetition of the process, 
the impurities can be concentrated at one end of the rod. This section is 
then removed and discarded. 

To grow single crystal n or p material, the purified semiconductor is 
melted in a crucible and the desired trivalent or pentavalent impurity added. 
A small seed crystal is dipped into the molten semiconductor and as it is 
slowly rotated and withdrawn, a large crystal grows on the seed with the 
same lattice orientation. This is illustrated in Fig. 1.6(a). 

The large crystals may be 2 or 3 cm in diameter. Properly oriented, they 
are sliced by diamond saws into thin wafers. The surfaces are then etched 
and polished to remove crystal dislocations caused by the sawing process. 



The Junction Diode 



11 




p Pellet 



Junction 



n Germanium 






i£ 




Molten 
Semiconductor 

(a) 



(b) 



(c) 



Figure 1.6 (a) Growing the single crystal ; (b) a grown pn junction ; (c) an alloyed 
junction. 

1.9 Manufacture of a pn Junction 

The semiconductor diode employs the properties of a junction between p 
and n layers of semiconductor material. In the grown junction process, a 
crystal is grown from an n melt and at a proper lime ap impurity is added. 
The p material overrides the n doping and yields a p material. The junction 
occurs where the n material changes to p, as indicated in Fig. 1.6(b). 

Alloyed junctions arc produced by fusing a pellet of a p impurity, such as 
indium, onto the w-silicon or H-germanium wafer. The indium alloys with 
the semiconductor and creates a thin /^-silicon or /7-germanium layer. The 
remainder of the pellet serves as a contact to the p side of the junction, as 
shown in Fig. 1.6(c). 

In the diffusion process, a semiconductor wafer of /(-impurity type is 
heated in a furnace with a gaseous atmosphere containing the desired p 
impurity. At elevated temperature the impurity atoms migrate into the 
semiconductor and create a p layer in the n wafer. Numbers of large wafers 
can be processed in one operation, resulting in uniformity of characteristics. 

There is no crystalline discontinuity at the junction formed between n 
and p layers in a single crystal. Such a junction cannot be created mechan- 
ically by forcing two individual crystals together since the junction must 
be atomic in nature. 



7. 10 The Junction Diode 

Our ability to make n and p materials by controlled impurity additions to 
a pure semiconductor results in the pn junction, which has the property of 
one-way conduction. Consider the junction diode of Fig. 1.7 with positive 
voltage connected to the p region and the negative battery terminal to the 



12 



The Semiconductor Diode 



The Diode Voltage-Current Equation 



13 
















Depletion region 

» ' T 1 n 
i u -r-| 1 






p " 




/ 


If 


++++ _l. 

+ +++ _ ; 

++++ , 


if 




•t 


+ + + + J ! | 

++++ ( i i 

i i i 


'» 














J 


\ J ! 


I.Ik 








ilil 




1 rM« 

+ V 








'■' + 

V 






(a) 








(b) 





Figure 1.7 (a) Forward voltage across a pn junction; (b) junction with a reverse 
voltage applied. 

n region. This is the forward current condition, with the forward current into 

p shown as l t . 

There are supplies of electrons in the n region supported by electron 
flow or injection into n from the negative battery terminal. There are holes 
in the p region supported by hole injection from the positive terminal. 
With the voltaee connected, holes move across the junction from p and 
into the electron-rich n region, where they meet and combine with electrons. 
But this positive charge has disturbed the neulrality of the n region and 
so electrons flow from the negative terminal toward the junction to restore 
the charge balance. This is indicated by the several flows of Fig. 1.8. 

Likewise, free electrons move from the electron-rich n region across the 
junction into the p region. This flow has made the p region more negative. 

We have a current composed of holes in the p region and a negative cur- 
rent due to electrons in the n region. The holes and electrons are continu- 
ously replenished from the battery and the movement of these charges to 
the terminals constitutes the forward current I, into the p and out of the n 

Figure 1 .8 shows how the number of charges varies with distance from the 
junction. The currents far from the junction are carried by the majority 
carriers of each region. Of course, in the metallic wires between diode and 
battery the current is due to motion of electrons. Electrons are supplied to 
the n region from the negative battery terminal and electrons are extracted 
from the covalent bonds at the ^-region terminal by the positive battery 
potential, in effect injecting holes into the p region. 

With reverse polarity, the diode has a positive voltage connected to the n 
region and negative voltage to the p region, as shown in Fig. 1.7(b). With the 
n region positive, free electrons are attracted away from the n junction. With 
the p region negative, the free holes are also pulled away from the junction 



and into the p region. As a result, there is a region extending between x, 
and .v 2 on both sides of the junction that has been depleted of mobile charges. 
Accordingly this is called the depletion region. It may be only a small fraction 
of a millimeter thick but without mobile charges it cannot conduct. Thus 
with reverse voltage across the diode we have an effective insulating layer 
at the junction. 

The depletion region is still subject to electron-hole pair generation, 
however, due to thermal energy. All such intrinsic charges arc swept out of 
the region by the applied voltage and this flow is known as the reverse satura- 
tion current I,. This reverse current depends on the number of broken cova- 
lent bonds at the operating temperature. The current is small in magnitude 
but it is strongly dependent on temperature. 

The pn diode easily passes current in the forward direction and has only 
a small current in the reverse direction. This is the property of a rectifier 
diode. 



Current 




Distance Along the Bar 
Figure 1.8 Electron and hole currents in a forward-biased junction. 



1.11 The Diode Voltage-Current 

Equation 

The current through a pn junction can be predicted by the equation 

/-JfcMT-l) (1.1) 

at 80°F or 27°C. The voltage V is positive for forward current and negative 
for reverse current. 

With forward voltages greater than 0.1 V we can use a simpler form of 






14 



The Semiconductor Diode 



equation as 

Forward Current 

I -UP" 0-2) 

and with reverse voltage more negative than —0.1 V we can use the relation 

Reverse Current 

I=-l, (1-3) 

This identifies the reverse current as the reverse saturation current of 
the junction. It is dependent on the rate of electron-hole pair generation in 
the depletion region and is not dependent on the reverse voltage up to the 
Zener (see Sec. 1.12) breakdown point, which occurs at about -40 V for the 
typical diode shown in Fig. 1 .9. The current /„ being dependent on the rate 
of pair generation, is variable with temperature. 











10 




\ 












o 

— — A 














/, m 


A 
















, i 




/ 






40 -. 


.f- 


20 


l 

10 


^L 


1 












* 


n o < in v 


1 




t 






-4jiA 
-8M 














1 





"tr 



fa) 



(b) 



Fifiure 1.9 (a) Typical semiconductor diode voll-ampcrc curve (note scale 
changes); (b) diode circuit symbol. 

Table 1.2 is calculated by use of Eq. 1.1 and presents the data in terms 
of the current ratio /,//,. It shows that the diode is a one-way electrical device. 

The volt-ampere curve of Fig. 1.9 is from actual measurement of a sample 
diode and the voltage includes resistance drops in the semiconductor material 
outside the junction region. This accounts for greater voltages than are given 
in Table 1.2. 

The diode circuit symbol is shown in Fig. 1.9(b). 



Zener Diodes 



15 



TABLE 1.2 Diode Forward Current 



+ V 



Ifll, 



I V 



IfU, 



0.04 


3.76 


0.06 


9.38 


0.08 


21.6 


0.10 


49.0 


0.12 


108 


0.14 


235 


0.16 


513 


0.18 


1119 


0.20 


2440 



0.22 


5.32 x 10' 


0.24 


11.6 x 103 


0.26 


25.3 x 10' 


0.28 


55.3 x 10' 


0.30 


120 x 10' 


0.35 


8.47 x 10' 


0.40 


5.96 x 10« 


0.45 


4.19 X 10? 


0.50 


2.94 x 10» 



1.12 Zener Diodes 

In theory, the reverse current of the pn diode is a constant, /„ and independent 
of reverse voltage. Actually the diode has a breakdown at some value of 
reverse voltage. This occurs at —40 V for the diode of Fig. 1.9. The break- 
down voltage is stable for a given diode. This breakdown is not harmful 
if the power dissipation in the diode is limited to the rated value and a safe 
temperature is maintained. 

The Zener diode circuit symbol is shown in Fig. 1.10(a). The diode 
voltage cannot exceed the breakdown value and V. at the output terminals 
will not rise above that level in the circuit of Fig. 1.10(b), regardless of in- 
creases in V,. The diode current increases and the excess of voltage V, over 
V, appears as an increased voltage drop in R. Therefore the Zener diode is 
useful to maintain a fixed voltage across its terminals as a voltage regulator. 
Its effect is much as if a battery V. was connected across the output terminals 
whenever V, > V„ as indicated in Fig. 1.10(c). 



+ 



If It 



6 



(a) 



/■ ' 



(b) 



K 



R 

O WvV- 

+ 



•& V ' 



(O 



Figure 1.10 (a) Zener diode symbol; (b) regulating circuit; (c) illustrating Zener 
circuit action. 



16 



The Semiconductor Diode 



The reverse-voltage region of operation was investigated by Zener and 
the diode bears his name. The reverse voltage appears across the depletion 
region and at some value is sufficient to accelerate the thermally generated 
electrons in the region to a velocity at which they are able to free more elec- 
tron-hole pairs by collision with covalent bonds. After each collision there 
is an additional pair of charges and in turn these arc accelerated and collide 
with atoms, creating still more charges; the current builds up like an ava- 
lanche. 

The breakdown voltage can be lowered by increasing the impurity dop- 
ing. Diodes are available with Zener voltages ranging from about 3 to 200 V 
and with power ratings from £ to 200 W. 



1. 13 Light-Emitting Diodes 

The gap energy E is acquired by the electron when it is broken free from a 
covalent bond in a semiconductor (see Section 1 .6). Consequently, when a 
free electron falls back into a hole in a covalent bond, the electron must 
give up the same amount of energy. This energy is radiated from the elec- 
tron as a single photon of light, with the color of the light related to the gap 
energy of the material. 

For a gallium arsenide pn diode with E a — l.5eV, the emitted light is 
a dark red. Green is obtained from a gallium phosphide diode. Because the 
energies of the electrons vary slightly, the light emitted is not of one wave- 
length but spreads over a narrow wavelength band. The intensity of the light 
can be varied with the applied voltage, changing the number of recombina- 
tions per second. 

The light is entirely electronic in nature and very fast on-ofF switching is 
possible. These light-emitting diodes (LEDs) are being used as point sources 
for numerical readouts for calculators and in alphabetic display devices. 



1.14 Photodiodes 



Reverse-voltage diodes in which the p layer is made so thin as to be trans- 
parent will conduct if light of proper wavelength falls on the junction layer. 
The photons received must carry sufficient energy to supply the gap energy 
and break a covalent bond; we then have a diode in which electron-hole 
pairs are generated in the depletion region by radiant luminous energy 
instead of thermal energy. The needed energy for silicon can be obtained 
with photons in the red and near infrared. An interesting application is 
made with the gallium arsenide LED, emitting in the red, and the silicon 
photodiode, sensitive in the red region. The two devices form a photoisolator. 



General Comments 



17 




(b) 

Figure 1.11 (a) The photoisolator, using an LED and a photodiode; (b) photo- 
diode symbol. 

as shown in Fig. 1.1 1. The two circuits are isolated from each other except 
for the light transmitted from the LED to the photodiode. There is no reverse 
transfer of energy. 

The reverse saturation current of the junction will still be present when 
the cell has no illumination and is known as the dark current. Its magnitude 
should be small compared to the current obtained with useful illumination 
levels. 



7. 15 Capacitance Effects in the pn Diode 

The pn junction with reverse voltage has a depletion region, acting as an 
insulator, between the n- and p-conductive regions. The combination acts 
as a capacitor. The depletion layer thickens with increased voltage as more 
charges are pulled away from the junction so that the capacitance has an 
inverse relation with applied voltage. 

Variation of the reverse voltage can be used to change the capacitance 
and ranges of 2 to 50 pF arc possible. These diodes are being used as tuning 
capacitors in high-frequency and TV receiving systems. They are known as 
varactors or voltage-variable capacitors (VVCs). 



1. 16 General Comments 

Our modern era of solid-state electronics relies on an understanding of atomic 
structure and electrical conduction processes. We employ the Bohr atom 
model with a positive nucleus surrounded by electrons in orbits, each having 
a fixed level of energy. Metals, as good conductors, have a great many free 
valence electrons to take part in electrical conduction. Conversely, in- 
sulators hold their charges very lightly and have very few free charges at usual 
temperatures; therefore, they do not readily conduct. The semiconductors 






18 



The Semiconductor Diode 



in the pure or intrinsic state have strong covalent bonding and as a result 
there arc few free charges available for conduction at room temperature. 
Impurity additions add free charges in controlled numbers as holes or 
electrons; with these tailor-made materials we are able to develop the one- 
way conduction properties of the pn junction. 

With forward bias the free majority charges are urged across the junc- 
tion and we have a condition of easy conduction. With reverse bias the 
mobile charges are withdrawn from the junction region leaving a high- 
resistance depiction zone. This acts as an insulator. 

The pn diode is a polarity-controlled switch, with remarkably low voltage 
required for forward conduction and a high resistance to conduction in the 
reverse direction. 

REVIEW QUESTIONS 

1.1 What are the parts of an atom? 

1.2 How docs a germanium atom differ from an atom of copper or an atom of 
argon ? 

1.3 How many electrons docs an atom of potassium have in its valence shell ? 

1.4 Why are silver and copper considered good conductors of electricity? 

1.5 What is meant by a free electron in a metal ? 

1.6 Describe covalent bonding. 

1.7 Describe the mechanism of conduction in intrinsic silicon at room temperature. 

1.8 At what temperature docs intrinsic conduction become limiting in silicon 
electronic devices? What is the equivalent temperature in germanium devices? 

1.9 What is a hole in a semiconductor? Why are there no holes in copper? 

1.10 Why is a single crystal structure desirable in semiconductor devices? 

1.11 What is accomplished in zone refining of semiconductors? 

1.12 What is the reason for adding a donor element to a semiconductor? Choose 
an element that might constitute a donor impurity. 

1.13 What is the reason for adding an acceptor element to a semiconductor? 
Choose a suitable clement for an acceptor impurity. 

1.14 What impurities might be used in the two sides of a pn junction ? 

1.15 Contrast n- and /?-silicon materials in several ways. 

1.16 Why is the permissible operating temperature of// silicon higher than it is for 
n germanium? 

1.17 Explain the development of the depletion region of a junction. 

1.18 What is meant by majority and minority charges in n silicon? What is the 
minority charge in p germanium? 

1.19 Explain how a hole moves through a p semiconductor, remembering that the 
impurity atoms are very far apart. 



Problems 



19 



1.20 Where are the minority carriers, and what are they, in a pn junction? 

1.21 At a given temperature, /, in germanium is larger than it is in silicon. Why? 

1.22 What kind of charges enter each end of the p region in a forward-biased pn 
junction? 

1.23 What is the effect of a forward bias on the depletion region in a pn junction ? 

1.24 What is the effect of increasing the reverse voltage on the depletion region of 
a pn junction? 

1.25 What is the cause of the reverse saturation current? 

1.26 Why does the reverse saturation current increase with temperature? How fast 
does ii increase in silicon? 

1.27 Why docs the reverse saturation current not vary with reverse voltage? 

1.28 What is a photon? 

1.29 What is the relation between the energies carried by a photon of red light with 
wavelength of 7.5 x 10 -7 m and a photon of ultraviolet light with wavelength 
of 3.75 x IO-'m? 

1.30 Explain the phenomenon of avalanching in a Zener diode. 

1.31 Explain how a varactor diode acts as a variable capacitance. 

1.32 Why is a pn junction sometimes said to be a nonlinear circuit element? 

1.33 Why is a diode called a polarity switch? 

1.34 The brightness of the LED is dependent on what factor? 

1.35 The color of an LED is determined by what factor? 

PROBLEMS 

1.1 Simplify the following quantities with use of the appropriate prefixes on the 
units: 

0.0001 m 9534 x 10"' F 

0.0000075 g 12,473 « 

1.75 x 10-s mm 0.00475 V 

0.0176 x 10"«s 0.0146 A 

1.2 What is the energy in electron volts of an electron that has risen through a 
potential of 3.25 V? 

1.3 For a silicon diode with /, = 25 /iA, find the forward current with a voltage 
of 0.2 V. 

1.4 A given diode has /, = 5 x I0~ 6 A; find forward and reverse currents for 
V = 0.25 V. 

1.5 A diode has /, = 5 x 10"' A ; what value of -I V is required to obtain a 75-m A 
forward current? 

1.6 A certain pn diode requires 0.28 V to cause a current of 180 mA in the forward 
direction. What is the reverse current at 35 V? 



20 



The Semiconductor Diode 



1.7 A Zcncr diode has a breakdown voltage of 20 V. The input to the circuit of 
Fig. 1 .10(b) is supplied from a 45-V battery and the series resistor R = 2500 Si. 
What should be the current and power rating of the Zener diode ? 

1.8 A Zener diode is rated at V, = 12 V and 6 W power. With V, = 15 V, find the 
needed series resistance R of Fig. 1.10(b) to keep the diode within power rating. 

1.9 A diode conducts 2 fit\ with 10 V applied and the same current with 150 V 
applied; is it forward- or reverse-biased in each case? What current will it carry 
when 0.2 V is applied in the forward direction? 

1.10 A pn diode carries 50 mA when 0.25 V is applied in the forward direction. 
What voltage is needed to raise the current to 150 mA? 



2 

The Diode as Rectifier 

and Switch 



The /JH-scmicondiictor diode acts as a switch whose on-off action is con- 
trolled by the polarity of the applied voltage. The switch has a low resistance 
and very small voltage drop under forward voltage and a very high resistance 
under reverse-voltage conditions. Numerous applications depend on 
this switching property, a major use being the rectification of alternating 
current for dc power supply. 

2. 1 The Ideal Diode Model 

The semiconductor diode has a voltage-current curve as in Fig. 2.1(a), pre- 
viously discussed in Sec. 1.11. The straight vertical line at Fig. 2.1(b) is an 
approximation to the actual curve and represents the ideal diode, assumed to 
be a closed circuit for a forward voltage and an open circuit for any reverse 
voltage. This seems a reasonable assumption since the usual forward voltage 
is less than 1 V and the reverse current is negligibly small at reverse voltages 
below the Zener level. In effect, we have said that a small forward voltage is 
required to make a diode conduct but we arc going to call that small voltage 
zero for ease in calculation. 

The diode electrodes are called the anode and the cathode. The anode is 
represented by the arrow of the diode symbol and is internally connected to 
the p region, while the cathode is the bar of the symbol and is connected to 
the n region. For forward conduction the anode should be positive and the 
cathode negative. 

21 



22 



The Diode as Rectifier and Switch 



- V 



_l I L 



1 2 3 + V 



- V 







+ V 



(a) (b) 

Figure 2.1 (a) pit diode volt-ampere curve; (b) ideal diode curve. 

2.2 The Half -Wave Rectifier Circuit 

On the half cycle of transformer voltage in which the diode anode is positive 
in Fig. 2.2(a), the diode acts as a closed switch. Voltage v of the transformer, 
Fig. 2.2(b), is then directly connected to the load and a current /' is present 
in the circuit as a pulse of half-sine form shown in Fig. 2.2(c). On the next 
half cycle of transformer voltage the diode anode is negative; the diode is 
in its reverse condition and acts as an open circuit. No current exists in the 
load. 



AC 



(a) 



^dc 





Figure 2.2 (a) Half-wave rectifier circuit; (b) transformer secondary voltage; 
(c) load-current pulses. 



The Half -Wave Rectifier Circuit 



23 



The successive half-sine pulses of Fig. 2.2(c) all lie above the zero axis and 
the waves have an average positive value, shown as I ic . The peak of the ac 
voltage from the transformer is V m and on the conducting half cycle this 
voltage appears across R. Then the peak of the current pulse is 



V I 41 K 
m R ' R 

With this definition we can determine the dc or average load current as 

/ _ h, - £k «. 031 8K m 
~ ItR —R— 



n 



(2.1) 



(2.2) 



and the dc voltage across the load is 

K dc -W dc /? = 0.318K„, 



(2.3) 



On the reverse half cycle the diode is an open circuit and the entire trans- 
former voltage appears across the diode. At the peak of the wave this is V m . 
The insulating depletion layer in the diode has to withstand the peak reverse 
voltage (PRV) in order to prevent a current. The PRV is an important rating 
of a diode. For the half-wave circuit 



PRV=K m = 1.41 V, n 



(2.4) 



The average current rating of the diode selected for a particular application 
must be greater than the expected I ic of Eq. 2.2 and the peak current rating 
of the diode must be greater than the expected peak current l m of Eq. 2. 1 . 

Example: A transformer has a rated secondary voltage of 24K rrai . Find 
the average current, the peak current, the dc voltage across the load resistance 
of 50 Q, and the needed PRV rating. 

From our knowledge of sine waves we know that 



V m = \A\V, m , = 1.41 X 24 = 33.8 V peak 



Then by Eq. 2.2 

By Eq. 2.1 

By Eq. 2.3 
Using Eq. 2.4 



/ --^ = ^ : 30- • 2, ^ 



L = if = ^r ~ omA peak 



Kic = /ic* - 0.216 X 50 = 10.8 V 
PRV= V m = 33.8 V peak 






The Diode as Rectitier and Switch 



The Bridge Rectifier Circuit 



25 



2.3 The Full- Wave Rectifier Circuit 

The full-wave rectifier circuit of Fig. 2.3(a) operates in a more efficient manner 
by supplying current to the load on both half cycles. The transformer is center- 
tapped and each half develops a peak voltage V,„. These windings arc op- 
positely connected to two diodes so that positive voltages are supplied to the 
diode anodes on alternate half cycles, Fig. 2.3(b), and they transmit current 
pulses to the load as shown in Fig. 2.3(c). 




Figure 2.3 (a) Full-wave rectifier circuit; (b) transformer voltage; (c) load- 
current pulses. 

In the first half cycle, diode Z), is forward-biased from 7, and conducts a 
current pulse to the load; diode D z has a reverse voltage from 7* 2 and appears 
as an open circuit. In the second half cycle, the transformer polarity reverses 
and diode D 2 is forward-biased from T 2 ; it transfers a current pulse to the 
load. At this lime diode D, is reverse-biased and open. 

With the transformer windings and diodes operating independently as two 
half-wave rectifiers, V m is the peak voltage of each half of the transformer 
to center tap. The peak current of a diode remains as 



" R 



(2.5) 



The load dc current is the average of two half-sine pulses, however, and is 
therefore twice the dc value obtained for the half-wave circuit, giving 



. _ 2(0.318K m ) _ 0.636K„ 
/dc R - —R— 



(2.6) 



The dc load voltage is 

V ic - I^R = 0.636K m (2.7) 

The two diodes can be seen as in series across the full secondary voltage 
of the transformer with a peak voltage of 2V m . One diode is a closed path and 
the other diode is open at any time. The voltage across the open diode at the 
peak is the PRV, which is 

PRV = 2V m (2.8) 

The full-wave rectifier circuit requires two diodes and a larger center- 
tapped transformer secondary, from which we obtain twice the dc voltage 
obtained from the half-wave circuit. However, the required PRV rating for 
the diodes also doubles. 

More importantly, the "bumps" of the load current pulses of Fig. 2.3(c) 
are a better approximation to a "smooth" and steady dc current and are 
easier to filter into smooth dc current. The use of the full-wave circuit is 
strongly favored over the half-wave circuit. 

2.4 The Bridge Rectifier Circuit 

At the small cost of two more diodes we can eliminate the center-tapped trans- 
former and obtain full-wave rectification in the bridge rectifier circuit of Fig. 
2.4(a). 

The transformer has one winding with a peak voltage of V m . On the first 
half cycle, with terminal A positive, diodes D x and D 3 have forward-voltage 
conditions and conduct in series as shown in the equivalent circuit at Fig. 
2.4(b). Diodes D z and D A are reverse-biased by the transformer voltage in 
this half cycle and are open circuits. A current pulse having I m = VJR passes 
downward in the load R. 

On the next half cycle the transformer voltage reverses and terminal B 
is positive; diodes D z and D 4 conduct, with the current path as in Fig. 2.4(c). 
Diodes D, and D } are reverse-biased by the transformer voltage and are open. 
Another current pulse passes downward in the load R. 

The current pulses in the load are in the same direction and the load 
current waveform is that of Fig. 2.3(c) as a full-wave output. The equations 
for dc current and voltage of Sec. 2.3 apply. 

Now consider point C in the circuit at Fig. 2.4(c). Point C is effectively 
connected to A of the transformer because the voltage drop across D z is 
negligible during conduction. With C at the voltage of A and point E con- 
nected to B, the open diode D 3 appears connected across the full transformer 
and has to withstand a peak voltage V m . The diode PRV rating therefore 
should be V m . Similar reasoning could be applied to each of the other diodes 
during the appropriate half cycles. 



26 



The Diode as Rectifier and Switch 




5 + 


c« 


Djw 




^0, 


I" 


or 


*\ 








E 













Mc 



(a) 




A» 




(b) 



(O 



Figure 2.4 (a) The bridge rectifier circuit; (b) with terminal A positive; (c) with 
terminal B positive. 

The bridge rectifier is extensively used because of the lower cost of the 
transformer and the lower PRV rating for the diodes compared to the center- 
tapped transformer circuit. 



2.5 Measurement of the Ripple 
in the Rectifier Circuit 

A battery provides a constant output voltage and is considered an ideal dc 
voltage supply. However, it is usually more economical to supply dc to 
electronic equipment from the ac line through rectifier circuits, with batteries 
used only for portable equipment. As has been shown, the output voltage 
from rectifier circuits is not steady, and has varying components or ripple 
superimposed on the average value of the dc voltage, as represented in Fig. 
2.5(a). Such variations, remaining from the ac supply as 60- or 120-Hz 
frequencies, will appear in an amplifier output as "hum" or noise. 

The amount of ripple in the rectifier circuit output is measured by the 



The Capacitor Filter 



27 



(2.9) 



ripple factor, y (gamma), defined as 

_ ripple voltage, rms 
* dc voltage 

Per cent ripple = 100 y 

Smaller values of y indicate smoother output voltage, or a closer approach to 
ideal dc. 

An ac voltmeter will read the rms value of the ac or ripple component 
when connected to the rectifier load through a large capacitor to block the 
dc voltage. The dc voltage can be measured by a dc voltmeter connected 
across the load, as in Fig. 2.5(b). 



Wl£ 



Peak 
Ripple 



(a) 



-W- 




Load 



I 



(b) 




Figure 2.5 (a) Ripple on a dc voltage; (b) circuit for measurement of ripple 
percentage. 

If such measurements are made on the output of the half-wave rectifier 
circuit, shown in Fig. 2.2(c), the ripple factor would be 1.21, or 121 percent; 
that is, the ac rms component in the output voltage is 1.21 times as large 
as the dc voltage. Similarly, the output of the full-wave circuit of Fig. 2.3(c) 
shows a ripple factor of 0.48 or 48 per cent of the dc voltage. 

Neither of these ripple figures is sufficiently low for use with most elec- 
tronic circuits where the expected ripple factor should be in the range of 
0.005 to 0.00005, or 0.5 to 0.005 per cent. The latter figure represents a ripple 
voltage of 0.5 mV superimposed on a dc voltage of 10 V. These requirements 
show that further smoothing of the rectifier circuit output is needed. It is best 
to start with full-wave rectification if filtering is to be added, since the original 
value of ripple is already lower than with the half-wave circuit. 



2.6 The Capacitor Filter 

The availability of capacitors of many microfarads makes the simple capacitor 
filter of Fig. 2.6(a) desirable as a low cost method of smoothing the rectifier 
output voltage wave. 



28 



The Diode as Rectifier and Switch 




(b) 

Figure 2.6 (a) Capacitor filler wiih a full-wave rectifier; (b) load and capacitor 
voltage. 



Figure 2.6(b) shows the capacitor and load voltage as v c , at the output 
of a full-wave rectifier circuit. From time 7", to the peak at nil the diode />, 
is forward-biased by the transformer sine voltage and the capacitor charges 
through Z), to the peak of the wave at V m . When the transformer voltage 
falls below V m after ji/2, the diode becomes reverse-biased and conduction 
stops. If no load resistance were connected to the circuit, the output voltage 
would remain at a dc level equal to V m . With a load R connected, the capacitor 
discharges through R and its voltage falls below V m until time T 2 is reached. 
At that time, diode D 2 becomes forward-biased by an excess of transformer 
voltage over v c and this diode recharges C to V m for the second half cycle. 

The fall in v c will be small during the discharge interval from »c/2 to T 2 
if C and R are large. This means that the ripple is small. The lime given to 
capacitor charging will be very short, possibly less than 10° of the cycle. We 
are able to develop an expression for the ripple of the capacitor filter with a 
full-wave rectifier as 



7 = 



0.144 
CRf 



(2.10) 



where /is the frequency of the supply line. This result introduces the term 
CRf as a useful design factor. 



The Capacitor Filter 



29 



What do we measure with the factor CRf"! We know that the period of the 
supply sine wave is T — l//and so 

CRf^^ 

Thus CRf represents the ratio of the capacitor-load resistance time constant 
to the period of the sine wave of the supply. In more general terms, CRf 
measures the ability of the capacitor C to store energy and maintain a voltage 
near V m until the next diode conduction period occurs to recharge the capac- 
itor. 

For low values of ripple factor we want CRf to be large. The result is 
indicated in Fig. 2.7, plotted for Eq. 2.10. Since the ripple varies inversely 
with the load resistance, then the ripple increases with load current for the 
capacitor filter. 




0.04 



0.03 



Q. 



0.01 



5 10 20 50 100 200 500 

CRf 

Figure 2.7 Full-wave rectifier: ripple and V ic \V m versus CRf for a capacitor 
filter. 



Given a specified ripple factor, we have need for 

0^44 p 
yRf 



(2.11) 



The dc load voltage is related to the peak of the ac voltage. By study of 
the wave form in Fig. 2.6(b), we have 



V* =V m -$A 






30 



The Diode as Rectifier and Switch 



where V R is the peak-io-pcak ripple voltage. The dc voltage is obtainable as 

1 






i 



i 



(2.12) 



ACRf 



This ratio again involves the design factor CRf and the equation is plotted 
in Fig. 2.7 against that factor. It shows that for no load (infinite R) the voltage 
will be equal to K m) as predicted. 

In Fig. 2.7 we see that CRf> 5 is needed for a ripple less than 0.03, or 
3 per cent. With CRf> 5 the dc voltage will be greater than 0.95 K,,. On this 
upper plateau of the voltage ratio curve, the load (value of R) may change 
considerably; yet the dc voltage will remain at 90 per cent or higher of the 
V m value. Thus the filter will give nearly constant dc voltage for varying loads 
as are encountered with some amplifiers. 

Example: A full-wave rectifier circuit, operating from a 60-Hz line, sup- 
plies a load with 200 mA and 30 V dc at full load. What value of C is needed 
to limit the ripple factor to 0.01 (I per cent ripple)? What should V m be? 

The ripple will be maximum at full load current, or minimum load resis- 
tance, which is 



R = 



30 

0.200 



= 150fi 



Then, using Eq. 2.1 1 we find C for the filter as 

0.144 0.144 



C = 



yRf 0.01 x 150 X 60 
- 0.00160 F- 1600 //F 

The ripple voltage is (30 X y) = 30 x 0.01 = 0.3K rra ,. 
The design factor is 

CRf- 0.0016 X 150 X 60 = 14.4 
Using Eq. 2.12, 



30 




1 


1 


v n 


-30x 


1 


1.017 


v m * 


■ X 14.4 
1.017 = 


30.5K peak 



and this is the needed transformer voltage. 

Figure 2.6(b) indicates that the pulse of current through the diode is of 
short duration but has a large peak value. In this short interval, the diode 
must pass the total charge which appears in the load as the average current 



The n Filter 



31 



over the half cycle. The peak current requirements of the diode are difficult 
to calculate because we do not know the length of the diode conduction 
period. Fortunately we are protected against design inaccuracies by the ability 
of the diode to withstand large current pulses for a short period of time and 
by the resistance and reactance introduced into the circuit by the transformer. 
At initial turn-on of the circuit, however, the filter capacitor is fully dis- 
charged. This capacitor acts as a short circuit on the transformer and con- 
ducting diode. The diode will have a surge current rating J s and the initial 
surge of charging current must be kept below this rating by use of a resistor 
in series between transformer and each diode. The value of this surge resistor 
can be calculated from 

*-* 

2.7 The n Filter 

For applications requiring ripple percentages much less than 1 per cent, the 
cascaded filtering action of the circuit in Fig. 2.8 is useful. The combination 
of C„ L, and C 2 is called a n filter because of its similarity to that Greek 
letter. 

I 

D\ I A = 



Filter and Load 



-D 



1 w 1 i~~T — 


-'000' » 

L 




+ 

fe V m sin u/ 

• 




r' 


H 






: + 

*V m sinw/ 






r R • 


















Di 




A 







I 



'dc 



Figure 2.8 Hull-wave rectifier with a n filter. 

Capacitor C, charges to the peak V m of the applied voltage as does the 
capacitor C of Sec. 2.6. Therefore we can use what we have already learned 
about the ripple voltage of the capacitor filter and determine the effect of 
/. and C 2 in reducing the ripple reaching the load. 

We shall consider only the full-wave rectifier circuit, as half-wave supply 
is rarely used with complex filter circuits. Accordingly, the ripple voltage has 
a frequency twice the line frequency. The effect of the inductor is due to its 
reactance at double line frequency: 

X L = 27r(2/I) = AnfL (£2) (2.13) 



32 



The Diode as Rectifier and Switch 



The effect of capacitor C 2 on the alternating ripple at the double frequency is 

1 1 



X c . = 



(ft) 



(2.14) 



c ' 2ti(2/C 2 ) 4tt/C 2 

Now C ; will be chosen sufficiently large in capacity that its reactance will 
be very small with respect to the load resistance R(R ^ I0AT c > for example). 
Let us calculate the parallel impedance of C\ and R as 



Using R = lO-Vc, we have 



JR 1 I X c , 






jmxi, i 



But yioi s v^oo « 10 so 

* ~ \ox c , ** 

The impedance of a parallel combination of a small capacitive reactance and 
a large resistance is effectively that of the capacitance only. For the ripple 
components of the current we have changed the actual circuit of Fig. 2.9(a) 
to the simpler one at (b). 



A 


-—— — ^ 












L 









k. 




Cr. 


r K ' : 





(a) 



A 

o- 







Mc 



(c) 

Figure 2.9 (a) The n filter ; (b) equivalent circuit for the double-frequency current ; 
(c) dc equivalent circuit. 



The n Filter 



33 



Equation 2.10 was written for the ripple factor across the capacitor filter 
and can be used for the ripple factor across C, of the n filter in Fig. 2.9, as 

0.144 



?■ = 



(2.15) 



C x Rf 

The ripple factor across C : may also be calculated. The voltage-divider action 
of L and C 2 further reduces the ripple voltage and we can write 

_ 0-144 Xc. 
71 C x RfX L + X Ct 

But we choose L so that X L > X c , and we can drop X c , in the denominator, 
giving 

0.144 X e . 
7z C,Rf X L 

Substituting for the reactances with Eq. 2. 1 3 and 2. 14 and using the values of 
the constants, we have 



(2.16) 



yi = 



9.1 x 10- 



(2.17) 



RC x CiLp 

The ripple factor is inversely proportional to the value of load R. Thus the 
ripple increases with load current. The inductance and the capacitances have 
an equal effect on the ripple so that choice of values and their distribution 
among C,, C 2 , and L is arbitrary. 

The dc voltage across C, at A, A is given by Eq. 2. 1 2 as 

V V - 

AA I - l/(4C,Rf) 

The inductor may have a dc resistance R c , and this resistance and the load R 
are connected in series across A, A for the dc current, as shown in Fig. 2.9(c), 
which is the equivalent circuit for dc. Using resistances R c and R as a voltage 
divider, we have the dc output voltage of the filter as 

For small load currents (large R) the dc output voltage of the filter approxi- 
mates V m if C, is large. Usually C, and C 2 are made large and equal, and L 
is given a value of 5 or 10 henrys. 

Example: Filter components might be chosen as C, = 50 /*F, C 2 = 
50 fxF, L - 5 H, and rt c - 200 £2, with a load taking 200 mA at 125 V dc. 
The load resistance is 

125 



/? = 



0.200 



= 625Q 









34 



The Diode as Recti/ier and Switch 



The Voltage-Doubling Rectifier Circuit 



35 



The ripple factor is 

0.00091 

72 ~ 625 x 50 x 10-" x 50 x 10" 6 x 5 x 60 3 
= 0.00054 

Per cent ripple = 100 y, « 0.054 per cent. 

This would be a satisfactorily small ripple factor for electronic amplifiers. 
It represents an mis ripple voltage of 

125 x 0.00054 = 0.067 V 

or 67 mV ripple on the 125-V dc supply. 

2.8 The n-R Filter 

For filters used with rectifiers having small current output, it is often econom- 
ically desirable to replace the inductor of a n filter with a resistor, as in Fig. 
2.10(a). 

The equivalent dc circuit of Fig. 2. 10(b) is the same as that for the n filter 
in Fig. 2.9(c) and so the dc output voltage can be calculated by use of Eq. 
2. 1 8, repeated here as 



V*c = 



1 + 



r 

4C,Rf 




A 



-WAr- 



K 5 '„. 



o- 
A 



(b) 



Figure 2.10 (a) The x-R filler circuit; (b) equivalent dc circuit. 



The resistance of the filter inductor R c usually is in the range of 80 to 400 Q 
and similar values arc suited to R f in the n-R filter circuit. 

The ripple factor is found by replacing X L = 4nfL, the reactance of the 
filter inductor, with R, of the filter circuit and so 

0.0114 ,, , n . 

Since the average power lost in R f is P = l} e R f , the circuit should be 
employed only when / dc is small to avoid undue power losses and the resul- 
tant heat dissipation in the equipment. 

Example: Replacing the 5-H inductor of the example of Sec. 2.7 with its 
resistance of 100 £2 as R f , we find the ripple factor becomes 

0.0114 

72 625 x 100 x 50 x 10 6 x 50 x 10" 6 x 60 2 
= 0.0202 ripple factor 
Per cent ripple = 100 y z — 2.02 per cent. 
This represents a ripple rms voltage of 

125 x 0.0202 = 2.53 V 

The ripple has been increased with the removal of the inductor. The dc 
voltage will be unchanged, however, since we made R, = R e . 

2.9 The Voltage-Doubling Rectifier 

Circuit 

A higher value of dc voltage can be obtained from a given transformer by 
use of the voltage-doubling rectifier circuit, one form of which is shown in 
Fig. 2.11. 

With the upper ac terminal positive we have the equivalent circuit of 
Fig. 2.11(b). With diode D, forward-biased and conducting, diode D z is open. 
Diode D, charges C, to the peak value of V m , the applied ac wave. On the 
next half cycle the circuit becomes that at Fig. 2.1 1(c), with B positive. Diode 
D 2 conducts through C 2 , charging that capacitor to the peak of the ac wave. 
The capacitor polarities are additive across the dc output terminals; at no 
load this voltage is equal to the double-peak value of the ac input wave. 

The diodes conduct high peak currents for short intervals and C, and C 2 
must be of high capacitance to maintain the dc voltage over the nonconducting 
intervals of the diodes and to give a low ripple percentage. 

The doubler circuit is useful for high dc voltages because the transformer 
has a lower total voltage and the transformer insulation requirements are 
reduced in comparison to other circuits. Performance data are presented 
in Fig. 2.12. 




Figure 2.1 1 (a) A voltage-doubler rectifier circuit; (b) and (c) equivalent circuits 
on alternate half-waves. 



2.0 



1.8 



1.6 

• *M 



1.4 



1.2 



- 


\ ' ' 


MM 




- 1 -i— 




- 


- 












- 


' / 


Ac 


\ Ripp 


le 






- 


/ 


i i 


.1111 




1 J— 


iTm 


- 



0.08 



0.06 






0.02 



10 20 

CRf 



50 100 200 



Figure 2.12 The voltage-doubler rectifier circuit performance. 



36 



Rectifying AC Voltmeters 



37 



The principles used in the voltage-doubling circuit may be extended to 
obtain still higher voltages. Such voltage multipliers use a ladder arrangement 
to increase the output dc voltage but applications are limited because of high 
ripple percentage with small output currents. 



2.10 Rectifying AC Voltmeters 

Instruments for the measurement of ac cannot employ iron in their field struc- 
tures because the action of the iron varies with frequency and such variation 
will introduce errors with varying frequency or waveform. With only air in 
the magnetic circuits, the ac instruments operate at low field-flux densities 
and must have large currents in the moving coils to give useful deflections. 
Permanent magnet-moving coil (PMMC) instruments for dc measurement 
derive their field flux from permanent magnets, however, and a smaller 
current can be measured than is possible with ac instruments. Therefore, 
we often combine diodes with PMMC instruments, rectifying the small ac 
currents and performing their measurement with low-current dc instruments. 
The bridge rectifier circuit of Fig. 2. 13(a) is used, leading to ac voltmeters 




o w- 



AC 



if 



(a) (b) 

Figure 2.13 (a) Full-wave average-reading meter; (b) peak-reading ac voltmeter. 

that may have internal resistances of 2000 £2 per V of scale or give full-scale 
deflection with a current of only 0.5 mA. The dc instrument coil is supplied 
with full-wave rectified current pulses and the deflection is proportional to 
the average or dc value of these pulses. From Eq. 2.6 for the full-wave rectifier 
circuit, 



/ dc = 0.6361* = 0.636 F 



+ Ro 



(2.21) 



38 



The Diode as Rectifier and Switch 



where 

V m = peak value of the applied ac voltage 

R m = series multiplying resistance of the instrument 

R =-■ the resistance of the instrument movement 

We assume that ideal diodes are being used. 
Inverting the equation above we have 



K m = 



0.636 



which is the peak value of the ac voltage applied to the instrument for a given 
dc instrument current. To determine the rms value of the ac voltage at the 
instrument terminals, we multiply by l/^/~T — 0.707 and have 



0.707K„= W&iR. 



RoV, 



Jc 



0.636' 
- l.\\(R„-R B )I ac (2.22) 

where /,,,. is the current measured by the deflecting coil of the dc instrument. 

The scale introduces the factor 1.1 1, which is appropriate only for a sine 
waveform from which we obtained the factors 0.636 = 2/n and 0.707 = 
\/*J~2 . If the input wave is not a sinusoid, the instrument indication will be 
in error. 

Another form of rectifying voltmeter appears in Fig. 2.12(b) and employs 
a half-wave diode with a capacitive filter. The voltage applied to R m and the 
instrument is equal to the peak of the ac wave. The scale is calibrated to read 
in rms values of that wave and includes the 0.707 factor. Again, errors arise 
if the applied waveform is not sinusoidal. 



2.11 Diode Wave Clippers 

The polarity-switching property of the diode permits its use in simple clipper 
circuits to remove parts of waveforms. The circuits employ a diode, a resistor, 
and often a battery or voltage source. The output waveform can be clipped 
at different levels, dependent on the diode connection and the battery polarity 
and potential. Basic forms include series and parallel diodes and biased and 
unbiased circuits. 

The circuit of Fig. 2.14(a) employs a parallel diode and scries resistor R, 
connected to clip off all positive voltages; essentially the circuit is a half- 
wave rectifier. When point A is positive to ground, the diode is forward- 
biased and appears as a short circuit, through use of the ideal diode model. 
For v, > 0, the state of the circuit is shown in Fig. 2.14(b) and v„ = 0. For 
v, < 0, the diode is reverse-biased and open and the state of the circuit be- 
comes that of Fig. 2.14(c). For a sine-wave input the output wave is that at 



Diode Wave Clippers 39 

Fig. 2.14(e). Resistor R should be large with respect to the diode resistance; 
10,000 Q is suitable for most diodes. 

A biased parallel clipper is shown in Fig. 2.15(a), with clipping of all posi- 
tive voltages above I 10 V. With v, < + 10, point A is negative to the diode 



n A A A r\ . 


A 


__o 


R 








"i 


W- 


O 











(a) 



-WW- 
R 



Vi>0 



« o =0 



(b) 



v,<0 



^vvW- 
R 



"o="l 



T 



(c) 








(d) 



2ir 




Figure 2.14 (a) Parallel diode positive clipper; (b) circuit of (a) for «/ > 0; 
(c) circuit of (a) for ut < 0; (d) input sine voltage; (e) clipped sine voltage. 



o \AAAi 




i 

h o 


R 








! :d 




+ 

i°:e 


°" 



o VWV 



A 



v,< 10 



vW— — » o 

R * 

Open 

v„ = 10 ■=- 



o W/v- 

R 



v, > 10 



'« !')■=- 



* 10 



(;■) 



(b) 



(cl 




(d) 



+ 10V 




(e) 



Figure 2.15 (a) Clipping at -f 10 V; (b) for v, < 10; (c) for * > 10; (d) and (e) 
waveforms. 



40 



The Diode as Rectifier and Switch 



cathode al -f 10 V, the diode is open, and the circuit is that of Fig. 2.15(b). 
The input voltage is transmitted to the output as ?•„. With v, > +10 V, the 
diode closes and the circuit becomes that at Fig. 2. 1 5(c), giving 1 V across 
the output. For sine-wave input the output wave appears at Fig. 2.15(e). 
In analysis of the circuit to determine the output waveform, the several 
circuits should be drawn for each voltage state, noting that switching between 



Scries 



o 1* 




*VW 



(a) 



*V. 




(b) 



Parallel 



""~innr 



(c) 



Figure 2.16 Clipper circuits, with sine-wave inputs. 



o vwv 

R 






— o 


o— 




Dy. 


— o 


O \MAr 

R 


[ 


— o 






»T 




"i 




h 


"o 


o— 




"J 


—o 


o VWV- 

R 


I 


— o 






/)? 




«/ 






"o 


o^ 




4, 


—o 



o VWV— • « 



:crmf 



(a) 



OAA 



/.. 'i 



(h) 



n n . 



'n 'i 



Figure 2.17 (a) Sine wave clipped lo form a square wave; (b) reshaping of dis- 
torted pulses by clipping. 



Diode Clampers 41 

the circuits occurs at V B in each case. Series and parallel forms of clipping 
circuits are compared in Fig. 2.16. Reversal of diode and battery polarities 
will rotate the waveforms about the zero voltage axis. 

Several applications of clipper circuits are shown in Fig. 2. 17. At (a) is a 
double-diode clipper that can be used to form approximate square waves from 
a sine-wave input. When followed by amplification and a second clipping 
operation the rise time of the pulses can be made quite short. At Fig. 2.17(b) 
we have the process of restoration of distorted pulses by clipping below the 
noise level. 



2. 12 Diode Clampers 

A clamping circuit employs a diode, a resistor, and a capacitor and will 
place a wave on a desired dc axis. For instance, Fig. 2.18(a) shows a circuit 
whereby the unsymmetrical input pulse at (b) is clamped with its positive 
peaks at zero in (c). 

The RC values of the circuit should be chosen so that the time constant is 
at least five times the period of the input wave; that is, with the period at \/f, 
then RC > 5/f. When the wave goes to +5V at t a , the capacitor charges 



-K- 



»< 



D v„ 



+ 5 


n 








(a) 








U 


'0 


'1 


'2 




u 


'0 


'1 


'2 




- 10 










- 15 












(b) 



(O 





-?r~ 


i 


+ 






5 ^ 


D 


"0 = 


1 








'0<'<'l 
(d) 



T_l 



10— 



D °a 



i>„= 15 



l\<K'2 
(e) 



Figure 2.18 (a) Diode clamp; (b) input pulse wave; (c) output clamped at V; 
(d) and (c) circuits during operation. 



42 



The Diode as Rectifier and Switch 



quickly through the diode to -I 5 V and we have the circuit condition of Fig. 
2. 18(c) that exists tor,. At/, the input drops to - 10 V, and with the capacitor 
charged to -f 5 V the diode voltage is —15 V and it opens, leading to the 
circuit at Fig. 2. 1 8(d) and - 1 5-V output. 

The output wave at Fig. 2.18(c) shows that the positive peak is now 
clamped aX V and the pulse swings down to —15 V. The positive peaks may 
be clamped at other dc levels by introducing a voltage V„ in series with the 
diode. 



2.13 Review 

The ideal diode model is useful in circuit analysis because of its simplicity; 
it employs the concept of zero forward-voltage drop when the diode anode is 
positive to the cathode and zero current with reversed polarity. An actual 
semiconductor diode provides a good approximation to the performance of 
the ideal diode in most applications. 

The full-wave rectifier circuit is an important application of the polarity- 
switching diode. The dc quantities in a resistance load are 



/dc = 



2V m . 



V*c = 



IV. 



The ripple, at 48 per cent of the dc voltage, is too large for a dc supply in most 
electronic circuits. 

The bridge rectifier circuit reduces the needed transformer voltage by one- 
half. This eliminates the relatively high cost of a center-tapped winding at 
the expense of adding two additional diodes. It is now a much-used circuit 
form. 

Ripple can be reduced by the addition of a filter capacitor across the load. 
The dc voltage will approach V m of the ac wave applied. The parameter CRf, 
with/as the supply frequency, is useful as a performance factor. The ripple 
reduction improves with increasing CRf values. 

Further reduction of ripple voltage is possible by use of the n filter but this 
adds an expensive and bulky inductor. Replacement of the inductor with a 
resistance saves cost, space, and weight but sacrifices dc voltage if the ripple 
is to be small. 

Filter circuits are compared in Table 2.1, several employ an input capaci- 
tor of many microfarads. This seems a better choice for modern small equip- 
ment than filter circuits using inductor input since the capacitor is lighter, 
cheaper, and smaller per unit of filtering effect than is the iron-cored inductor. 
A capacitor input filter also provides output voltages near V m and gives good 
control of that voltage if sufficient filter capacitance is used (large CRf). 
Thus these circuits make efficient use of the transformer. 









Review Questions 



43 



TABLE 2.1 Filter-Circuit Comparison (Full-Wave Rectifier) 



>"dc 



Ripple 



No filter 
Capacitor filter 

n filler 

n-R filler 



2K. 
n 



I I 



4CRJ 
I \R I R c ) 



4C,Rf 
K 



1 + 



\R I R f ) 



0.485 

0.144 
TRf 

0.00091 
RC,C z Lf> 

0.0114 
XR/CCtf* 



REVIEW QUESTIONS 

2.1 What is an ideal diode? 

2.2 A sine wave of voltage is applied to an ideal diode. At what angles in the wave 
does the diode switch on and off? 

2.3 The secondary voltage or a transformer for a half-wave circuit is 35K Im ,. What 
should the voltage rating of a transformer for a full-wave center-tap rectifier be, 
assuming equal dc voltage output from the rectifier? 

2.4 Define 

(a) Peak reverse voltage 

(b) Average current 

(c) Maximum current 

(d) Ripple 

2.5 Name several advantages of full-wave rectification over half-wave rectification. 

2.6 For rectifiers with resistance loads and 60-Hz supply, what is the frequency of 
the ripple component in the load voltage of 

(a) Half-wave rectifier 

(b) Center-tap circuit 

(c) A bridge rectifier 

2.7 What is the effect of a defective diode that is continuously open on the output 
voltage and the ripple of a full-wave rectifier? 

2.8 What are several advantages of the bridge rectifier circuit over the full-wave 
center-tap circuit? 

2.9 Name at least one disadvantage of the bridge rectifier over the full-wave center- 
tap circuit. 

2.10 What is the purpose of a filter circuit? How does it accomplish this purpose? 

2.11 Why do diodes carry short current pulses when a capacitor filter is used? 






44 



The Diode as Rectifier and Switch 



2.12 What components of the diode current pass through the load in a capacitor 
filter circuit? 

2.13 Why does the voltage output of a capacitor filter approach V n in magnitude? 

2.14 Name an advantage of a n filter over a capacitor filter. Name a disadvantage. 

2.15 What is the effect of the resistance of the inductor winding in a n filter? 

2.16 Name several disadvantages of filters employing inductors. 

2.17 Under what conditions do we use a n-R filter? 

2.18 What is the relation of the maximum voltage rating for the filter capacitors to 
the rms voltage of the transformer with a bridge rectifier circuit? 

2.19 What is an advantage of a voltage-doubler circuit ? 

2.20 Is the voltage-doubler circuit of Fig. 2.11 a half-wave or a full-wave circuit? 

2.21 If a PMMC instrument measures the load voltage of a half-wave rectifier, what 
is the component measured ? 

2.22 The diodes of the bridge rectifier of Fig. 2.13 are not ideal. How would you 
alter Eq. 2.22 to include the effective diode resistance? 

2.23 A dc voltmeter reads 680 V across the output of a voltage-doubler rectifier, 
at no load. What would an rms ac voltmeter read across the transformer 
secondary that supplies the rectifier? 



PROBLEMS 

2.1 Determine the dc voltage obtained by half-wave rectification of an ac wave of 
180-V rms. 

2.2 What PRV rating should the diode of Problem 2.1 have? 

2.3 A half-wave rectifier circuit supplies 100 mA dc to a 500 Q load. Find the dc 
load voltage, the PRV rating for the diode, and the rms rated voltage of the 
transformer supplying the rectifier. 

2.4 Find the dc load voltage obtained from a full-wave rectifier in which the trans- 
former supplies a peak voltage of 150 V on each side of the center tap. 

2.5 The dc output voltage of a full-wave rectifier is 120 V. What is the needed rms 
voltage rating for each half of the center-tapped transformer? What is the peak 
ac voltage rating for the full transformer secondary? 

2.6 A center-tapped transformer supplies the diodes of a full-wave rectifier giving 
a dc output across the load of 180 V. What PRV rating should be used for 
the diodes? 

2.7 Determine the needed PRV rating for a diode in a bridge rectifier supplying 
80 V dc to a load. 

2.8 A full-wave rectifier has 125 V dc output to a resistance load. What is the rms 
value of the ripple voltage? 

2.9 Calculate the needed transformer voltages to supply a load at 50 V dc, from 
(a) A half-wave rectifier circuit 



Problems 



45 



(b) A center-tapped full-wave circuit 

(c) A bridge rectifier circuit 

(d) Determine the diode PRV ratings for each case. 

2.10 A rectifier output following a filter has an rms ripple voltage of 1.72 V, with 
a dc load voltage of 90 V. What is the ripple factor? 

2.1 1 A bridge rectifier supplies a resistance load with 2 A dc at 20 V dc. What trans- 
former voltage is needed (rms)? 

2.12 A full-wave rectifier is supplied with an rms voltage of 80 V, 60 Hz on each side 
of the transformer center tap. A 10-//F capacitor is used as a filter and the 
load takes 50 mA dc. What dc voltage is being obtained at the load? 

2.13 A transformer with 250-0-250 V rms, 60- Hz secondary supplies a full-wave 
rectifier having a load of 1500 CI. Determine the dc load voltage and current, 
the PRV rating needed for the diodes, and the average current of each diode. 
Draw the circuit. 

2.14 A transformer with 250 V rms, 60-Hz secondary supplies a bridge rectifier 
having a dc load of 1500 SI. Determine the dc load voltage and current, the 
PRV rating of the diodes, and the average current in each diode. 

2.15 A silicon diode is used in a half-wave rectifier circuit operating from a 20 V rms, 
60-Hz transformer. The load is a 10-O. resistance. What is the peak diode cur- 
rent? What is the average load current? What should the PRV rating of the 
diode be? What is the rms value of the ripple voltage? 

2.16 A full-wave rectifier is supplied at 60 Hz by a transformer with a center tap. 
With a 400-fiF filter capacitor, calculate 

(a) The ripple 

(b) The rms ripple voltage 

(c) The rms voltage rating each side of center tap for the transformer, for a 
load taking 400 mA at 50 V dc. 

2.17 Operating from a 60-Hz line, a transformer supplies a peak voltage of 140 V 
each side of the center tap to a full-wave rectifier. The capacitor filter has a dc 
output voltage of 120 V and a dc current of 50 mA. Calculate the value of C 
used in the filter. 

2.18 An RC filter supplies 80 V dc with 1.5 per cent ripple. What is the rms ripple 
voltage? 

2.19 A rectifier transformer operates at 60 Hz and supplies a full-wave rectifier with 
a capacitor filter of 100 //F. The load takes 200 mA at 50 V dc. 

(a) What is the no-load (R very large) output voltage? 

(b) What should be the rms voltage rating of the transformer, measured to 
center tap? 

(c) What is the per cent of ripple at full load? 

2.20 A 60-Hz transformer having a 40-0-40 V rms secondary supplies a full-wave 
rectifier circuit. The load is 100 Q. with a filter capacitor of 3000 //F. Find 

(a) The dc load voltage 

(b) The rms ripple voltage across the load 

(c) The PRV rating for the diodes. 



46 



The Diode as Rectifier and Switch 



2.21 A transformer on a 60-Hz line supplies a bridge rectifier. A capacitor filter and 
a load of 24 SI gives a ripple of 1 .2 per cent at a dc voltage of 40 V. 

(a) What rms transformer voltage is needed ? 

(b) What capacitance is needed for the filter? 

(c) What should be the voltage rating for the filter capacitance? 

2.22 With a full-wave rectifier at 60 Hz, a n filter supplies 154 V dc to a load. With 
an ac voltmeter having a large series capacitor, the voltage measured across 
the load is 127 mV rms. 

(a) What is the ripple factor? 

(b) If the dc load current is 0.250 A, what value of C= C, — Ci is being used 
in the filter? L 10 H. 

2.23 Determine the ripple factor for a bridge rectifier circuit at 60 Hz with a n filter 
having C, C : - 100 fiF, L = 7.5 H, and a load current of 500 mA at 60 V 
dc. 

2.24 Determine the ripple factor for a bridge rectifier circuit at 60 Hz with a capac- 
itor filter of 2000 //F and a load taking 500 mA at 6 V dc. 

2.25 The circuit of Fig. 2.19 uses a transformer with 120-0-120 V im , secondary. 

(a) What is the magnitude and polarity of voltage at A with no load? 

(b) Find the same quantities at B, also with no load. 

(c) If the load at Sis 1000 £2, what is the ripple percent there, 60-Hz supply? 




Figure 2.19 

2.26 What is the dc voltage available at no load from a voltagc-doubler circuit 
supplied by a source of 120 V, 60 Hz, at no load? 

2.27 A bridge rectifier operating at 60 Hz is to supply 1 A at 35 V, with a ripple less 
than 0.5 V rms. Specify C and the transformer voltage rating, the average diode 
current rating, and the PRV rating. 

2.28 In Problem 2.27, what surge protection resistor should be used if the diode 
surge current rating is 30 times the average current? 



Problems 



47 



2.29 A bridge rectifier with n filter uses two 40-//F capacitors and one 10-H inductor 
of zero resistance. The transformer voltage is 300 V rms, 60 Hz and the dc load 
current is 150 mA. Find the dc output voltage and the ripple. 

2.30 The waveform of Fig. 2.20 is applied to the ammeter shown at Fig. 2.20(b), 
where M is a PMMC average reading instrument with 1-mA full-scale reading. 
If the diode is ideal, what does the meter read if its resistance is 9000 Q. 



+ l(K) V 



-« — ►- 
10 nu 



50 V 



50 

ins 



10 ins 



100 ms 



(a) 



(b) 
Figure 2.20 



3 

Models for Circuits 



Transistors and tubes arc called active devices because they convert energy 
from dc to ac at signal frequencies. Simple models are needed for the 
active devices, that can be used in circuit analysis along with resistors, capac- 
itors, and inductors. With such models we can design our electronic 
systems on paper and determine how ihey should be operated for best 
performance. Breadboarding and checking of the circuits can follow in 
the laboratory. Without preliminary calculation with models, ihe laboratory 
work becomes an exercise in frustration of the cut-and-try variety. 

Fortunately, those who work with electrical circuits have shown us how 
to develop models for circuits that contain energy sources, and are active. 
While seemingly rather abstract, we can use these methods to replace actual 
active devices with a simple circuit model that performs in an equivalent 
manner. 



3. 1 The Black Box Concept 

Consider a device as simple as a relay. It has an input pair of terminals or 
an input port and an output pair of terminals or an output port. If we apply 
a proper voltage (or signal) to the input port, we discover that a connection 
appears across the output port. If we know what happens at the output port 
for every input condition, we do not need to know the internal design to 
utilize the relay in a control system. The relay box might be painted black 

48 



The Black Box Concept 



49 



to hide its internal details, with just the ports exposed. Then the box of Fig. 
3.1(a) might represent the relay as a two-port network. A simpler one-port 
network is shown at Fig. 3.1(b). 



lo- 



lo- 



Nelwork 



-02 



-02 




(a) (b) 

Figure 3.1 (a) A two-pon network ; (b) one-port network. 

A great many of our electrical devices can be thought of as "black boxes" 
with one or two ports. We are able to apply them because of our knowledge 
of the electrical performance at the ports; we do not have to know what 
happens internally. For instance, we buy transistors by calling for a type 
number that assures us that the output will respond to the input in a certain 
way. We need not ask how the transistor is built internally; thus we buy 
transistors as black boxes. • 

We can study the performance of the one-port box of Fig. 3.2(a) by 
means of the current and voltage at the port. The values obtained indicate 
the response of the box to a connected source or load. Now, consider the 
second box in Fig. 3.2(b). If for every value of applied voltage K, = V z we 



I o- 



io- 



Network 
A 



I o- 



1 °- 



Network 
B 



Figure 3.2 Two one-port electrical networks. 



find that current /, ■■ / 2 , we would conclude that the electrical networks 
inside the boxes were the same in external performance. One box could be 
replaced with the other without noticing any difference at the ports and we 
would say that the boxes are equivalent. We need not know anything about 
the internal operation or connections of the boxes. We conclude 

that a circuit is equivalent to a second circuit if the second circuit can 
be substituted for the first without change in the currents and voltages 
appearing at the ports. 

Now, if we can devise a circuit model that operates at input and output 
ports the same as does a transistor, then we can use that circuit model in 
place of the actual transistor. 



50 



Models lor Circuits 



I 



In the networks of Fig. 3.1, the conventional current and voltage polar- 
ities are indicated. The currents moving inward are made positive at both 
ports of the two port so that a network can be reversed or changed end-for- 
end without altering an analysis. 



3.2 Active One-Port Models : 
the Voltage-Source Circuit 

Suppose that we connect a voltmeter at the port of Fig. 3.3(a), and discover 
that a voltage K, is present. We conclude that the box contains a source of 
electrical energy and houses an active network. It would be useful to deter- 
mine a circuit equivalent for this box. 



|o- 

+ 



lc~ 



Active 
Network 




(») (b) 

Figure 3.3 (a) An active one-port network; (b) the series form of equivalent 
circuit. 

We know that some internal voltage source is present and we speculate 
that the source might have some associated impedance. We then have two 
unknowns, a source V and an impedance Z', that can only be connected in 
two ways— in series or in parallel. We might start with assumption of the 
series form at Fig. 3.3(b). We need two algebraic equations to determine 
the two unknown quantities; these equations might be obtained by connect- 
ing two arbitrary loads at the port and measuring the resultant voltage and 
current. 

Probably the simplest loads we can find are an open circuit and a short 
circuit; the latter will cause no damage as long as we work only on paper! 
We measure the voltage at the port of Fig. 3.3(a) without load and call that 
value V, = V,,,.. Using the circuit of Fig. 3.3(b) and an open-circuit load, 
we measure V at the port. For equivalence of the two boxes we must have 

V,. = V (3.1) 

Now apply a short circuit to port 1,1 in Fig. 3.3(a) and we measure a 
current I, = I ]C ; similarly with a short circuit on the port of (b) we measure 
Ii = I'|. For equivalence of the two boxes our second condition is 

I.c - r, (3.2) 



Active One-Port Models: the Current-Source Circuit 



51 



With the short circuit in place across the l,l terminals of Fig. 3.3(b), 
we can write a circuit equation 

V ZT - 
Substitution of the necessary conditions for equivalence as given by Eq. 3. 1 
and 3.2, we have 

V M Z'I, C - 

and 

(3.3) 



z 77 



Therefore we have in Fig. 3.3(b) an equivalent circuit for the box at (a) if 

V - V oc (3.4) 

(3.5) 



Z'=^c 



The equivalent circuit includes a voltage generator V equal to the open- 
circuit voltage at the port of the original network and a series impedance Z' 
determined from the open-circuit voltage and the short circuit current. This 
circuit in Fig. 3.3(b) can be made equivalent to any one-port network and is 
known as the voltage-source equivalent circuit or the Thevenin equivalent 
circuit. 

3.3 Active One-Port Models: 
the Current-Source Circuit 

Let us place a load Z L at the output port of the voltage-source equivalent 
circuit in Fig. 3.4(a). A voltage equation written around the loop is 

V - Z'\ L - \ L = (3.6) 

Division by Z' and rearrangement of the terms gives 

V'_, . Vl 
T ~ h + Z 





(a) 



(b) 



Figure 3.4 (a)Thcvcnin's scries equivalent circuit and load; (b) Norton's parallel 
equivalent circuit and load. 



62 



Models lor Circuits 



But V'/Z' = I,. by Eq. 3.4 and 3.5 and we can call V t /Z' = I,, so that the 
equation above becomes 

I* - h + Ia (3.7) 

This equation is that of a current summation at point A for the circuit in 
Fig. 3.4(b). The circuit in the box consists of a current generator I„ and a 
parallel impedance Z', as derived for the voltage-source circuit. 

The circuit in Fig. 3.4(b) was derived from the voltage-source circuit at 
(a) and so we have another active one-port equivalent circuit, the current- 
source equivalent or the Norton equivalent circuit. By mathematics we have 
found the parallel form of circuit that we originally stated as possible. 

Since the voltage-source circuit and the current-source circuit are equiva- 
lent, transference from one form to the other is often used in circuit sim- 
plification. 











(a) 



(b) 
Figure 3.5 Circuits for the example. 



Example: We find that with open circuit at 1,1 of a box the voltage is 
5.2 V. With a load of 2 Q connected to those terminals the current is 1.7 A. 
Determine the circuit values for active one-port equivalent circuits. 

Since we have two unknowns in the box, we need two equations for solu- 
tion. From the problem data, 

V M - V - 5.2 V 

is one relation. The second relation comes from the equivalent circuit in 
Fig. 3.4(a) as 

V - IT -IZ L = 
With / given as 1 .7 A for Z L -= 2 £2 and V = 5.2 V, we have 

5.2- 1.7Z' - 1.7 x 2- 

1.7Z' = 5.2-3.4= 1.8 

z' = |^= i.06 n 

Therefore our voltage source should be that in Fig. 3.5(a), with V" = 5.2 V 
andZ' - 1.06 SI. 



53 



Maximum Power Output 

We can calculate 7, c as 

^ = F = m- 4 - 9IA 

and the current-source circuit is that in Fig. 3.5(b) with I K = 4.91 A and 
Z'= 1.06 ft. 

3.4 Maximum Power Output 

Many of our electronic power sources, such as microphones, are expensive 
and develop very little power output. For these generators we try to maxi- 
mize the power output per unit of investment by operating such sources 
under conditions that will give the maximum power output to a load. 

The voltage source in Fig. 3.6(a) represents any active one-port circuit; 
for simplicity we assume that the load is resistive and Z' — R'. If we use an 
open-circuit load at 1,1, the output current will be zero and there will be 
zero power output. Likewise, if we use a short-circuit load at 1,1, the output 
resistance is zero and there will be zero power output. These are the limits 




(a) 



0.5 

Im 

B 

§ 

O 0.3 

b 

B 

E 

& 0.2 


0.1 0.2 0.5 1.0 2 5 10 

R L /R' 
(b) 

figure 3.6 (a) A voltage source and load R L : (b) output power to the load versus 
RlIR: 





— i — i — 


i>-n 




i i 


i i i i 


























r 










"J 




i i 


i i : i 




l. _i_ 


i ii i. 



54 



Models lor Circuits 



for the load and if we experimentally try other load values, the curve in 
Fig. 3.6(b) develops. This curve shows that the maximum power is delivered 
from the active generator circuit at 1,1 when 

R L - R' (3.8) 

This is the condition of a matched resistive loud. 

With the same current in R L and R\ the load power is P L — l l R L and the 
internal power loss is P a = PR'. With R L = /?', these powers are maximum 
and equal. The power transferred to the load is only 50 per cent of the total 
power generated, however, or the output efficiency is 50 per cent. Needless 
to say, industrial and domestic power systems cannot afford to waste one- 
half of their output in the generating plant and do not operate with matched 
loads; such systems use loads much higher than the matching value to achieve 
high output efficiency. 

Example: For the voltage-source equivalent circuit in Fig. 3.5(a) the 
current / is 

R' I R L 

and for the matched load we have R' — R L so that 

5.2 5.2 



W 2 X 1.06 



2.12 



= 2.45 A 



The power delivered to the matched load is 

P L -- PR L - 2.45 2 X 1.06 = 6.37 W 
The total power generated is 

P G - P(R' - R L ) = 2.45* X 2.12 = 12.73 W 
and the efficiency of delivery of power to the load is 50 per cent. 



3.5 The Two-Port Network 

We also make frequent use of a transmission type of network in which there 
arc input and output pairs of terminals, as for the two-port network sym- 
bolized by the black box in Fig. 3.7(a). There are four undetermined quan- 
tities at the ports of this network, namely V u /,, V 2 , and l 2 . Consequently, 
if we are to find an equivalent circuit for any network in the box, we shall 
need four measured parameters. 

We can relate four parameters with four terminal voltages and currents 
by a pair of equations, and electronic usage has shown these desirable: 

V, = h,I, + h,V 1 (3.9) 

h = Vi + h - y i (3.10) 



The Two -Port Network 



55 



1 o- 



I o- 





(a) 



Qv. 



ii v-\ 



<b> 



-o2 



hZ v * 



-o2 



Figure 3.7 (a) A two-port black box; (b) the //-parameter equivalent circuit. 



where these are alternating voltages and currents. The subscripts of the h 
parameters are a standard usage. 

Borrowing from the method already used for the one-port circuit, we 
can take measurements at the ports with open- and short-circuit loads and 
choose the internal elements of the equivalent circuit to yield the same ter- 
minal measurements. 

With a short circuit at 2,2 we have V 2 — and, inserting that condition 
in Eq. 3.9 and 3.10, we find 

V t - v, 

h - v. 

and so we are able to define /;, and h, by the following measurements made 
at the 1,1 terminals: 



*-a 



.•2 ihorl 



= short-circuit input resistance 



(3.11) 



h f = -f-\ = short-circuit forward current gain (3.12) 

' I J2.2 Ihotl 

Taking measurements at the 2,2 port with an open circuit at the 1,1 
port means that /, = and using that condition in Eq. 3.9 and 3.10, we have 

v, = hjr, 
h - V. 

Then we define It, and h by the following measurements made at the 2,2 



56 



Models for Circuits 



terminals: 

V 1 
h, = Tr\ — open-circuit reverse voltage gain (3.13) 

" 2 Jl. 1 open 

h„ = rr = open-circuit reciprocal output resistance (3.14) 

' aJli I open 

The defined // parameters include an input resistance, the reciprocal of 
the output resistance, and two dimensionless ratios: consequently, the /; 
coefficients are called the hybrid parameters. 

With alternating voltages and currents the parameters /?, and //„ should 
be stated in units of impedance and reciprocal impedance but they are usu- 
ally considered as resistive. 

3.6 The h-Parameter Equivalent Circuit 

With h parameters corresponding to the measurements, Eq. 3.9 and 3.10 
will act as a mathematical equivalent for any two-port network. Our remain- 
ing step is to find an actual circuit that can be used to represent that original 
box. 

The hybrid equations were set up as 

V t - V, + h,V z (3.15) 

h = V. + h V z (3.16) 

Equation 3.15 appears as a sum of voltage terms and, with K, at the port, 
the right-hand terms represent the sum of voltages inside the box. The h,I, 
term is a resistance voltage drop and is shown as due to //, at port 1,1 in 
Fig. 3.7(b). The second term represents a controlled voltage generator, whose 
output is dependent on the voltage at the output port. The parameter h, is 
the constant of proportionality. Equation 3.15 is simulated by the series 
circuit at port 1 . 

Equation 3.16 states the sum of currents at A in Fig. 3.7(b). Current / 2 
enters at the port and /i K 2 leaves through resistance \/h . The second term 
is a leaving current due to a controlled current generator, whose output cur- 
rent is proportional to the input current at 1,1; Ay is the constant of propor- 
tionality. We have the circuit connected to port 2 representing Eq. 3.16. 

Figure 3.7(b) is an equivalent circuit for any two-port network, defined 
in h parameters. It should be noted that it comprises two one-port equiva- 
lents, with a voltage-source equivalent circuit appearing in the input loop 
and a current-source circuit appearing in the output loop. 

Example: Find the // parameters for the circuit in Fig. 3.8(a), and draw 
the equivalent circuit. 

A short circuit at 2,2 connects the 12- and I5-Q resistors in parallel and 



The h-Parameter Equivalent Circuit 



67 



1 n— 


i 

i 


Sft A 12ft 


1 


— o •> 


1 v^- 

+ 


I 








+ 


*i 


i 
I 

i 

! 

1 
1 
I 




.1512 




. -o 7 


i °— 


1 







v i. 



(a) 



ly — 



l°- 



L. 



•II.7J2 



^lp.5S6 V 2 

0.556 f> ' 



-02 



0.037 
7J 



(b) 
Figure 3.8 Circuits for example. 

the input resistance at 1,1 is due to 5 Q plus the parallel value, that is, 

(s.c.) fc-^-S + gf^-HJO 

With the short circuit at 2,2, current /, divides at A. Then 

12 x 15 



y M - /. 



Also 

We can then write 



12+ 15 



_/ _ y M ... i »S 

2 12 ~ ' 12 -F 15 



(s.c.) *, = £= -^=-0.556 



With an open circuit at 1,1 there is no current in the 5-Q resistor and so 
the output resistance measured at 2,2 is 

r = 12+ 15 = 27 Q. 

(o.c.) A. = — = i= = 0.037 mho 
To 2.1 



68 



Models lor Circuits 



Still with an open circuit at 1,1, the voltage across the I5-Q resistor is V t . 
Then 



'*"" 12 | 15 
V, = 15/, = 



15K, 



12+ 15 



and so 



(o.c.) h, = £ = £ = 0.556 



The circuit parameters are entered on the circuit diagram in Fig. 3.8(b). 



3.7 Power in Decibels 



In the communications field it is usual to measure power gain in decibels. 
Devised by the telephone industry and originally named the bel for Alexander 
Graham Bell, the decibel is one-tenth as large and of a more useful magnitude. 
The decibel (dB) is defined as 



number of dB - 10 log^ 
* i 



(3.17) 



where log indicates a logarithm to the base 10. 

The ear hears sound intensities on a logarithmic intensity scale and since 
many amplifiers supply an audio output, the use of a logarithmic power scale 
becomes reasonable and convenient. In fact, it was intended that a 1-dB step 
be the minimum intensity change that is just noticeable by the human ear 
but experiment has found that the minimum noticeable intensity change is 
nearer 2.5 dB. 

To illustrate the use of the decibel, assume the output of an amplifier 
under one condition is 3.5 W with an input of 0.12 W. The power gain, 
measured in decibels, is 

dB= 10 log ^ = 10 log 29. 17 

= 10 X 1.46 = 14.6-dB power gain 
If the amplifier output is increased to 7.0 W, then the decibel gain is 

dB= I0logi^= 10 log 58.3 

= 10 x 1.76 = 17.6-dB power gain 

Doubling of the power gain is shown by a 3-d B change. 
An amplifier with a power gain of 90 has 

dB- 10 logy -10 x 1.95- 19.5 dB 



Power in Decibels 



59 



and the amplifier has a gain of 19.5 dB. It is followed by a second amplifier 
with a power gain of 250 and 

dB - 10 lqgyL= 10 x 2.40 = 24dB 

The overall power gain is 

90 x 250 = 22,500 

and in decibels this is 

dB= 10 log 22,500- 10 log (2.25 x I0 4 ) 
= 10 (log 2.25 I 4)- 10(0.352 • 4) 
= 43.5 

But 43.5 ■= 19.5 — 24 and we see that the overall gain of amplifiers in cascade 
can be calculated by adding the respective gains in decibels as 

overall gain, dB - dB, + dB, + dB 3 + . . . (3.18) 

While defined as a power ratio, the decibel is also used for absolute power 
measurement when a reference or zero level is stated for P x . A variety of zero 
levels has been used; the one now commonly employed is 0.00 1 W (1 mW). 
Consequently we can state the output level of the amplifier with 7.0- W output 
as 

7 



10 log 



0.001 



= 10 log (7 x 10 3 ) 

- 10(log7 + 3) = 10(0.84 + 3) 

- 38.4 dB above 1 mW 



This unit is sometimes given as dBm (decibel referred to one milliwatt), to 
indicate the 1-mW zero level. 

The amplifier with 3.5-W output has decibel output 



10 log 



3.5 
0.001 



- 10 log (3.5 x I0 3 ) 



= 10(0.54 + 3) - 35.4 dB above 1 mW 

Taking the difference of the 7.0- W output and the 3.5-W output, 

38.4 - 35.4 = 3.0 dB 

which was the original difference in the amplifier outputs. If this figure had 
been negative, a loss would have been indicated. 

We may wish to find the dBm of a microphone with 0.0000062-W out- 
put. With zero level at 0.001 W, we write 

dBm - 10 log kffjffi 62 - 10 log 0.0062 



Obtaining the logarithm of a number less than unity need pose no problems. 



so 



Models for Circuits 



Problems 



61 






Using scientific notation we can employ our regular methods as 

dBm = 10 log (6.2 x 10 3 ) = 10 (log 6.2 - 3) 
- 10(0.792-3)= -22.1 

and the result is staled as 22.1 dB below 1 mW. 

Power is given by P = V 2 jR or PR. The decibel power gain can be found 
through use of cither of these expressions. For instance, 



If R z = R„ then 



But we have 



gain in dB- lOlogp^ 
gain in dB = 10 log (j^V 



and we find that the gain can be written 



gain in dB - 20 log ^ 



(3.19) 



We commonly calculate amplifier gains by use of Eq. 3.19 as a voltage ratio 
or as the current ratio 



gain in dB - 20 log ^ 
'i 



(3.20) 



3.8 Comments 



The concept of an equivalent circuit, equal in terminal voltages and currents 
to any actual network of unknown internal elements and connections, is a 
very useful means of simplifying our frequently complex electrical circuits. 
The voltage source (Thevenin circuit) and the current source (Norton circuit) 
serve as equivalents for one-port active networks and, as controlled sources, 
have already appeared as parts of the //-parameter two-port equivalent cir- 
cuit. 

The latter circuit will be employed to represent the transistor or tube in 
circuit analysis. With the proper values for the h parameters, we can represent 
the transistor as a circuit of resistances, capacitances, and generators and can 
write and solve circuit equations that demonstrate the amplifying effect of 
the transistor. 

REVIEW QUESTIONS 

3.1 What is a "black box" intended to represent? 

3.2 What do we need to know about the internal connections of a black box? 

3.3 Why can we say that a voltage-source circuit is equivalent to a black box of 
one port? 



3.4 What are the two measurements needed to determine the circuit elements of 
a voltage-source equivalent circuit? 

3.5 How arc the circuit elements of a voltage-source circuit related to a current- 
source circuit ? 

3.6 If the one-port box contains no energy source or is passive, draw the equivalent 
circuit. 

3.7 Why are we interested in obtaining the maximum power output from some 
generators? Name two such sources. 

3.8 What is meant by matching a load? 

3.9 What is the output power efficiency of a generator with a matched load? 

3.10 What kind of terminal loads are used in determination of the h parameters 
for a two-port equivalent circuit? 

3.11 Why are the assumed currents directed inward at the ports of a two-port net- 
work? 

3.12 Sometimes the use of a short circuit leads to generator damage. Can you sug- 
gest an alternative means of obtaining data to calculate the // parameters with- 
out using a short circuit? 

3.13 Why are the h parameters called hybrid parameters? 

3.14 What is the major property of a controlled source? 

3.15 Why arc h, and h f called control parameters? 

3.16 What is one advantage of rating amplifiers in decibels of gain? 

3.17 What is the usual reference level when we state that an amplifier output is at 

130 dB? 

3.18 What is the meaning of a circuit input of -30 dBm? 

3.19 Determine the logarithm of 0.00672. 

3.20 Can you suggest why h, is negative to //, in the example in Sec. 3.6? 

3.21 Why is the decibel well suited to measurements along telephone lines? 

3.22 The noise level of a jet engine is measured at | 137 dB. What is the decibel 
noise level of 2 jet engines? Of 10 engines? 

PROBLEMS 

3.1 In Fig. 3.9(a) we find V, - 107 V at open circuit; with a load of 200 £2 we find 
K, = 49 V. Determine the circuit elements for a voltage-source equivalent 
circuit. 

3.2 Determine a current-source equivalent circuit for the circuit of Problem 3.1. 

3.3 Under short circuit, Fig. 3.9(a) has an output current of 8.7 mA. With a load 
of 10,000 Q we find the port 1,1 voltage to be 57 V. Draw and label the circuit 
elements for a voltage-source equivalent circuit. 

3.4 

(a) What is the load that should be used for maximum power output with 
the circuit of Problem 3.3? 

(b) What is the maximum power output available from the 1,1 terminals? 



62 



Models for Circuits 



Active 
Network 



-Ol 



-oi 




~L 



-ol 



ion 



-oi 



(a) 



(b) 



Figure 3.9 



3.5 What load should be used at the 1,1 port to obtain maximum power output 
from the circuit in Fig. 3.9(b)? 

3.6 Determine the element values for a current-source circuit equivalent to the 
circuit in Fig. 3.9(b). 

3.7 By use of the basic definitions applied to the circuit in Fig. 3.10(a), calculate 
the h parameters. 

3.8 Calculate the h parameters for the two-port circuit in Fig. 3.10(b). Draw the 
equivalent circuit. 

3.9 A generator of I V is connected to 1,1 in Fig. 3.10(b). Draw an equivalent 
voltage-source circuit at the 2,2 port and label the circuit elements with their 
values. 



I o- 



20 n 
-WA- 



15 £2 



-o2 



lo- 



lon 
-WA/- 



-02 



35 n: 



:sn 



I2fi 



lo- 



-o2 



I o- 



(a) 



(b) 



-o2 



Figure 3.10 



3.10 Draw and label the circuit elements for a voltage-source equivalent circuit for 
the circuit in Fig. 3.11(a). 

3.11 Transform the circuit in Fig. 3.11(a) into a current-source equivalent. Draw 
the circuit. 

3.12 What is the matching load for the circuit in Fig. 3.1 1(a)? 

3.13 Determine the voltage-source equivalent circuit for the circuit in Fig. 3.11(b) 
at the 1,1 port. What is the matching load? What is the maximum available 
power output to a load? 

3.14 Determine the current-source equivalent circuit for Fig. 3.11(c). Draw the 
equivalent voltage-source circuit. 






Problems 



63 



ion 




ion 

|o VWV- 



+ 
IOV-=- 



sn 

-VWv- 



lo- 



+ 

15 V-=- 



(b) 



4n 

o VvVv 1 




3 A 



(c) 



Figure 3.11 



3.15 A microphone has an output at -56 dBm. It supplies an amplifier that is to 
have its output at + 37-dBm level. 

(a) What is the amplifier gain required in decibels? 

(b) What is the amplifier power output in watts? 

(c) What is the amplifier input in watts? 

3.16 An amplifier takes 5 mW input power and has an output of 32 W. 

(a) What gain is present in the amplifier in decibels? 

(b) What is the dBm level of the output? 

(c) What is the dBm level of the input? 

3.17 An amplifier of 34-dB gain is connected in cascade with an amplifier of 28-dB 
gain. If 1 /iW is the input to the first amplifier, what is the output power in 
watts from the second amplifier? 

3.18 A radio receiver has an input resistance of 70 fi. The antenna supplies a cur- 
rent of 0.2 /iA to that input. The electrical power to the loudspeaker is 12 W. 
Find 

(a) The decibel gain of the receiver 

(b) The decibel level of the input signal, 1-mW zero level 

3.19 A transistor is found to have h, = 28 CI, h, = 4 x 10~ 4 , h t - 0.98, and h 
= 0.6 X I0 -6 mho. In a circuit equivalent to that in Fig. 3.7(b), find V 1 if /, 

1.5 //A. Him: Use the /i-parameter equations. 



4 

Junction Transistors 
as Amplifiers 



In 1948 Bardeen and Brattain found that the current through a forward- 
biased semiconductor junction could control the current to a reverse-biased 
contact mounted nearby. The result was the first solid-state control device and 
was called a transistor, as a contraction of the words transfer resistor. Such a 
transistor employs both holes and electrons in the conduction process and is 
said to be bipolar. Shockley later developed the field-effect transistor that 
employs only holes or electrons; it is therefore called a unipolar device. This 
will be discussed in Chapter 6. 

4. 1 Transistor Voltage and 
Current Designations 

Agreement on abbreviations used for transistor voltages and currents is 
needed for accurate understanding of transistor circuits. We shall use lower- 
case letters to designate time-varying quantities and capital letters for dc 
quantities and for rms values of ac signal voltages and currents. Illustrated in 
Fig. 4. 1 are some of the variations to be expected; these include 



Quantity 



Example 



Toial current or vollage /' c , v B 

AC signal, rms value l c , V b 

DC value l c , V c 

Supply or bias V cc , V B b 



64 



The Junction Transistor 



66 





r\ 












I V 










/ 










c 


/ 










V 


/ 










fa 


/ , 










3 


/ ' 










U 


J 










ha 












o 


4 






















u 












— 
73 






\ / '"' 




u 


>c 


>c 











• 











Figure 4.1 



Time 
Transistor current notation. 



The subscripts c or C, e or E, and b or B are used to identify the transistor 
internal elements; the collector, the emitter, and the base, respectively. 



4.2 The Junction Transistor 

Junction transistors are three-layer sandwiches of n and p materials, as il- 
lustrated in Fig. 4.2. These transistors are made in two types, dependent on 
the arrangement of the materials us pnp or npn. Performance of the two types 
is similar. In one the majority carriers are holes and in the other they are 
electrons. The bias voltages arc also reversed in the two types. 

Junctions are present at the interfaces between the n and/) materials. The 
three layers are designated as emitter, base, and collector. The first junction 
between emitter and base is operated with forward bias. The second junction 
between base and collector is operated with reverse bias. For the pnp unit the 
base and collector are maintained negative to the emitter; while for the npn 
unit the base and collector are positive to the emitter, as shown in Fig. 4.2. 

The pnp unit will be used for discussion of transistor operation. The emit- 
ter material is heavily doped in comparison to the base and, with forward 
bias, holes move from the p emitter to the hole-poor n-base region. With a 
very thin base, in the range of a few thousandths of a millimeter, the holes 
arc attracted by the negative collector voltage. The reverse bias on the second 
junction makes that junction one of easy flow for holes from the n-base 
region and the holes are swept across the junction to the collector. Thus we 
have established a current of holes from the emitter to the collector. Positive 
holes are shown by the arrows as inward at the emitter and outward at the 
collector terminal. 

The base width is made very narrow compared to the average distance the 
free holes move before recombination occurs in an n material. Therefore, 
while a few holes meet electrons and rccombinc in the base, 90 per cent or 



66 



Junction Transistors as Amplifiers 



Emitter Base Collector 




Emitter Base Collector 

. .\ A, . /. 




Figure 4.2 (a) Junction transistor, pnp; (b) transistor, npn; (c) circuit symbols. 

more of the injected holes will travel across the base and reach the collector. 
A small flow of electrons will occur in the base lead, to resupply the negative 
charge lost to recombination in the base. This flow creates an outward base 
current, as shown by the arrow in Fig. 4.2(b). 

The hole current from the emitter to the base is varied by the forward 
voltage across the junction, in accordance with V in the diode equation of 
Section 1.11. Control of the emitter-base voltage by a signal voltage causes a 
signal variation in the hole flow to the base and a similar variation in the col- 
lector current. The transistor is a control element in which the emitter-base 
voltage controls the collector output current. 

The input power is low because of the low resistance of the forward- 
biased emitter-base junction, typically a few hundred ohms. The output cir- 
cuit carries the same current, however, but typically will deliver it to loads of 
thousands of ohms. It is thus possible to have power amplification with the 
transistor. 



The Volt-Ampere Curves of a Transistor 



67 



The circuit symbols in Fig. 4.2(c) show the forward direction of emitter- 
base current by the arrow on the emitter element; the direction of the arrow 
distinguishes between pnp and npn transistors. 

The npn transistor could be discussed in like manner by reversing the bias 
potentials to maintain forward- and reverse-biased junctions and by consider- 
ing the emitter current as composed of electrons. The arrows on the transistor 
of Fig. 4.2(b) indicate the directions of forward current. 



4.3 The Reverse Saturation Current 

in Transistors 

The reverse-biased base-collector junction of the transistor is to be isolated by 
opening the emitter connection, as in Fig. 4.3. The main current of holes to 
the collector is removed but there still exists the usual saturation current of a 
reverse-biased junction, due to thermal pairs generated in the materials. 
Holes generated in the base material are swept to the collector by the negative 
voltage there and electrons generated in the collector are swept to the base. 




Figure 4.3 Condition for Icbo- 

The resultant current can be measured with the emitter circuit open as I CBO 
(meaning current, collector-to-base, emitter open). Current I CBO is always 
present with or without emitter current and is additive to the major i c com- 
ponent. Current I CBO is highly temperature sensitive and the operating tem- 
perature must be limited to keep I CBO small with respect to the desired control 
current component coming from the emitter. 



4.4 The Volt-Ampere Curves 
of a Transistor 



The output curves, Fig. 4.4(b), are the fundamental means of explaining the 
control characteristics of a transistor. These are curves of actual currents and 
related voltages and permit us to predict transistor performance. They will 



68 



Junction Transistors as Amplifiers 




i c <mA) 



fa 

-.1 — ■ L 




30 



10 20 

»CE 

(a) (b) 

Figure 4.4 (a) Transistor circuit; (b) output curves of a silicon transistor at 
25°C (77°F). 

provide a basis for the circuit models that will simulate the action of the 
transistor in amplifier analysis. 

The slope of the curves for any i B value is given by 

A/ c 



Av CE 

and dimcnsionaliy this is the reciprocal of a resistance. Accordingly, we can 
measure the collector resistance of the transistor by the reciprocal of the slope 
as 



r c — 



Av c 

Mr 



(4.1) 



The low slope of the curves shows that r c is high ; it appears relatively constant 
for a given base current but becomes lower for increased values of base cur- 
rent. 

At values of collector-emitter voltage of 1 V or less the curves merge into 
the saturation line, at the left in Fig. 4.4(b). Control of current by the base is 
lost. The reciprocal of the slope is again a resistance, that of the collector- 
base junction as a diode. This minimum resistance of the collector-base junc- 
tion is known as R CE i,.t)- It is largely due to the material resistance in the 
collector region. 

The collector current is controlled by variation of the base current, which 
is, in turn, determined by the base-emitter voltage v Ht: . In the region bounded 






The Current Amplification Factors 



69 



by the saturation line, the line for i„ — 50 fiA., v c£ = 25 V, and the line for 
i B = 0, the characteristics are uniformly spaced and regular, the slopes are 
relatively constant, and this implies constancy of the various transistor param- 
eters. This bounded region will be the region of operation for this transistor 
in many types of amplifiers. 

Cutoff of the transistor occurs for i H = 0, which is also the line for / c = 
Iceo (current, collector-to-cmitter, base open). For silicon transistors this 
current is very small, typically 2 ft A, and the curve for /„ = will lie very 
close to the abscissa. For a germanium unit the typical I CEO value might be 
100 to 200 ft A. 

40 



30 



< 
3 



20 



in 









1 

50°C 




\ 












1 


lo°c 










1 














/ i 


1 






































J 


1 










r-r^ 


/ 









0.2 0.4 0.6 0.8 1.0 



1.4 



Figure 4.5 Input characteristics for the silicon transistor of Fig. 4.4. 

The input curve of the transistor appears in Fig. 4.5. This shows the rela- 
tion between base current and base-emitter voltage in the forward-biased 
junction. This curve looks similar to a forward diode curve, which it is. There 
is considerable variation of v BK with temperature, as indicated. 



4.5 The Current Amplification Factors 

In Fig. 4.2 we showed the currents that were physically present at the transis- 
tor terminals; they reversed when we changed from a pnp transistor to an 
npn transistor. To avoid continuing confusion and to allow us to calculate for 
a circuit and then use either type of transistor, we are going to define all three 
transistor currents arbitrarily as positive inward, as shown in Fig. 4.6. That is, 

i E + ic + 'a = (4.2) 



70 



Junction Transistors as Ampliliers 



Relations between the Amplification Factors 



71 




'£ 

-o 






"BE 



k 



(a) (b) 

Figure 4.6 Defined currcnis and voltages: (a) pnp transistor; (b) npn transistor. 

and obviously at least one current must be negative and reversed to satisfy 
KirchhofTs current law. 

We now can develop several useful transistor parameters. The relation 
between collector and emitter currents is a measure of the efficiency with 
which the charges are transported through the base. A change in emitter cur- 
rent will differ from the resultant collector current change by reason of the 
recombinations occurring in the base. We measure the efficiency of transport 
at constant collector-base voltage and state it as the ratio of a small change in 
collector current A/ c - to the small change in emitter current A/ £ . The param- 
eter is useful for small changes or small signals and is called the small-signal 
collector-emitter current amplification factor. This has a magnitude of 



Al£_L„ CO 



(4.3) 



Recombination current is kept small through use of a thin base layer and 
the value of a lies in the range of 0.90 to 0.99. 

A good approximation to a can be obtained by using the steady current 
values at a point on the output curves; that is, we can employ 



BS&-A 



! « 



(4.4) 



as the static or tic collector-emitter current amplification factor. 

A second amplification factor relates a small change in collector current 
A/ c caused by a change in base current Ai B . This is determined at a con- 
stant collector-emitter voltage and is designated as the small-signal col- 
lector-base current amplification factor, written as 



h - A 'r] 
h " ~ ATj.. 



(4.5) 



Jpcfc'conitint 

Since i B is small by design, the factor /»,,'" ranges from 20 to 200. 

Again, we can approximate h f , by using the steady currents at a point on 
the output curves; that is, 

ki-kssk* (4.6) 

'// 

"Also called /? in some literature but that symbol will not be used here because of 
conflict with the notation used in feedback amplifiers. 



Equation 3.12, with i c as an output and i„ as an input value, will explain 
the choice of h f as a symbol for this amplification factor or current ratio. 

Example: Ati> cfi = 15 V and i B = 20 mA, the curve in Fig. 4.4 shows 
i c - 2.6 mA. Using Eq. 4.2, 

l E + i c + '* = 
i e + 2.6 + 0.02 = 

i t: - 2.62 mA 

Consequently, Eq. 4.4 shows for the approximate magnitude of a that 



H _ I 2.60 
"-232 



= 0.992 



at point Q on the characteristics. 



Example: The constant value of v c -e requires that changes be made along 
a vertical line in Fig. 4.4. Using v CE = 15 V, and with changes from the point 
at Q, with i B = 20 //A down to i B = 10 //A, the corresponding change in i c 
is from 2.6 to 1.4 mA. Then, 

h (2-6- l.4)mA . 1.2 _ „ 

" ~ (0.02 - 0.0 1 ) m A _ 0.0 1 " 

If we use the dc value as an approximation, we have at the same point Q 

''-^abTmA^ 130 ^^' 



4.6 Relations between the Amplification 

Factors 

The algebraic sum of the currents is equal to zero by Eq. 4.2 and so 

A/* + A/ c + At, -- (4.7) 

Equation 4.3 was written for the magnitude of a but in Fig. 4.6 we can see 
that i E and l c are oppositely directed so that we write 



A/. . A A- 



(4.8) 



Inserting this definition in Eq. 4.7, 



_A^ 
a 


+ A/ c - 


-M B 


A/ C (l 


-*)- 


— A/ fl 




A/ c 


a 




1 -a 



(4.9) 



72 



Junction Transistors as Amplifiers 



The Load Line and O Point 



73 



We recognize the left side of the equation as the definition of h { , and so we 
may write 



ft/.- 



(4.10) 



(4.11) 



1 -a 

Equation 4.10 may be rearranged to give 

1 + A,. 

Both Eq. 4.10 and 4.1 1 will be very useful in later discussions. Another use- 
ful equation can be obtained from Eq. 4.10 by writing 

(1 - a)A„ = a 
Inserting Eq. 4. 1 1 and canceling h fl leads to 

1 



1 -a = 



1 + A, 



(4.12) 



4.7 The Load Line and Q Point 

A simple transistor amplifier appears in Fig. 4.7. Writing a voltage equation 
around the output loop gives 

V CB = Vcc ~ Rk (4.13) 

This is the equation of a straight line.The transistor output curve family also 
involves the variables v CE and i c ; by plotting Eq. 4.13 on the family of transis- 
tor curves, the dc load line obtained will locate all possible values of v CE and 
i c for the series circuit of transistor and load R at a given V^. 




Figure 4.7 A transistor amplifier. 

To draw the load line for Eq. 4.13 we use the axis intercepts. Insertion of 
i c = in Eq. 4. 1 3 gives an .r-axis intercept at x = V cc ; use of v CB = gives 
>-axis intercept my = V CC IR. In Fig. 4.8 a load line is drawn for V cc — 20 V, 
V CC IR = 40 mA, and R - 20/0.04 - 500 Q. 




Figure 4.8 Output curves of a silicon transistor at 25 C (77°F). 

The load line is the locus for all voltages and currents that can exist with 
the load and transistor in series. If alternating input signals are used as shown, 
a zero axis point or no-signal point Q must be chosen on the load line. The Q 
point is also called the quiescent point since the circuit is quiet or V, = 0. We 
rather arbitrarily choose a Q point and decide on the bias currents and volt- 
ages, the steady V CE , the steady collector current I c , and the base current I B . 
We use these biases to place the signal variations in a region of the curves 
where the transistor will operate in most linear fashion. The bias sources also 
supply the dc energy that the transistor converts to the ac signal. Circuits for 
establishing the bias currents and voltages will be discussed in Chapter 5. 

For true reproduction of the input sine current signal, the output ac 
wave must be derived from equal /„ excursions along the load line, each side 
of the Q point. As the signal voltage moves positive, the operating point 
moves along the load line up to A at the positive peak, back to Q, and down 
to B at the negative peak of the i B signal input. The output current i c varies 
correspondingly, as shown along the i c axis in Fig. 4.8. 




74 



Junction Transistors as Amplifiers 



Having selected a Q point, we can determine the h parameters of the 
transistor in the region around that point. Wc shall then use those parameters 
in the equivalent circuit of the transistor. The value of h fc can be found from 
a measurement along a vertical line in Fig. 4.8 at constant v CE . The change in 
i c per unit change in i B represents h f , and, for example, a change of /j from 
200 to 150 ^A causes a change in i c from 16.2 to 1 1.7 mA. Then 



. _ (16.2 - 11.7)mA .. on 
" (0.20- 0.15) mA 



(0.20-0.15) 
The output short-circuit reciprocal resistance is defined as 

and this represents the slope of a constant i B curve in Fig. 4.8; it is apparent 
that h oc is constant over the central region. 
The input resistance is defined as 

h V, Av BK 
h "-T i --Ei7 

This relation represents the reciprocal slope of the input curve in Fig. 4.5, as 
an example. The input signal variation about the Q point must be kept small 
for //,, to be considered constant. The requirement for constant values of the 
h parameters is the reason that h fr and the oiher h parameters arc defined as 
small-signal parameters. 

The parameter h„ can be found as the slope of a curve relating v CE and 
v Bt: for constant i c . 

4.8 The Basic Transistor Amplifiers 

The transistor is an active amplifier element with three external leads, one 
connecting to the emitter, one to the base, and one to the collector. We shall 
use it in circuits as a two-port device with four terminals. As such, one of the 
transistor leads must be made common to the input and output ports. The 
choice is arbitrary and leads to the three connections in Fig. 4.9. These are the 
common-emitter (C-E) circuit, the common-base (C-B) circuit, and the com- 
mon-collector (C-C) circuit. 

These three circuits differ in performance and application. For example, 
we shall find that the C-E circuit is best for power gain. The C-B circuit is a 
load-matching device that permits a low-resistance source to be efficiently 
connected to a high-resistance output circuit and results in a voltage gain 
near unity. The C-C circuit reverses the resistance transformation property 
and gives a current gain near unity. Other differences in the characteristics of 
these three circuits will become apparent with further study. 

To indicate the common input-output electrode and the use of the C-E, 



Simplification of the Equivalent C-E Circuit 



C-E 



C-B 



C-C 




75 



tt-O 



4/" \'< T " Vee 1 

o — i- * o o 1- 




(a) 




(b) 

Figure 4.9 Defined currents in C-E, C-B, and C-C connections with (a) pnp 
transistors; (b) npn transistors. 

C-B, or C-C amplifier circuit, a second subscript is employed with the h 
parameters, as h f „ h lb , h oc . In this text we intend to use only the h parameters 
for the common-emitter configuration; however, conversion equations for 
all three configurations are presented in Sec. 4.14. 



4.9 Simplification of the Equivalent 

C-E Circuit 

Any linear two-port circuit can be replaced with the //-parameter circuit of 
Chapter 3. We have just shown how the transistor can be connected as a two- 
port circuit element. Therefore, we can replace the C-E connected transistor 
in Fig. 4.10(a) with the /i-parameter circuit in Fig. 4. 10(b). The values of the/; 
parameters can be taken from the manufacturer's data or from the charac- 
teristic curves but greater accuracy will result if they are measured at the 
selected Q point. 

A simple common-emitter amplifier is drawn in Fig. 4.1 1(a), including the 
bias voltages V BB and V cc that establish the Q point. In Fig. 4.1 1(b) we have 



76 



Junction Transistors as Amplifiers 




h t 



B 



'c 



V. 



0* 



d'b 



■ir I K, 



(a) 



Transistor 



(b) 



Figure 4.10 (a) The C-E transistor as a two-port clement; (b) the /i-parameter 
equivalent circuit as a two-port element. 




\h*>2 |o- 



B >' 



o2 lo- 



\ C 



h„y„\ 



"ff'h 




/~ 

Transistor 
(a) (b) 

Figure 4.11 (a) C-E transistor amplifier; (b) A-paramcter equivalent for (a). 

replaced the transistor with its equivalent circuit between B, C, and E. Since 
no ac signal voltage can appear across the dc bias sources, they are eliminated 
from the circuit. The output blocking capacitor is chosen so large that no ac 
signal voltage appears there and it too may be removed from the equivalent 
circuit. Later we shall consider its effects. With voltages V, and V„ as rms 
values, we now have the complete ac equivalent circuit shown in Fig. 4.1 1(b). 
The //-parameter circuit equations of Sec. 3.6 apply to this transistor 
amplifier between ports 1,1 and 2,2. Using voltage and current designations 
from Fig. 4. 1 1 we have 

V, = h„I b + h r .V (4.14) 

/, - h ft I„ + K,V (4.15) 

Because of the direction of I c and the assigned polarity of the voltage across 
the load, we must write 



*-- 5 



(4.16) 



The Transconductance g„ 



Tl 



Using this relation in Eq. 4.15, we have 

— ^=h f .h + h M V. 

Solving for the input current 1„ and inserting the result in Eq. 4.14, we have 
an equation involving V, and V as variables. We can write the equation for 
the voltage gain as 

-1 



A " ~ V, ~ 



/>„(! +h„R) 
h„R 



-K, 



(4.17) 



The performance of the amplifier can be better understood if we can find 
a less complicated expression for the voltage gain. Fortunately it is possible to 
simplify the circuit and the resultant equations with small error. 

Generator h„V c is a control source that feeds back some of the output 
voltage to the input circuit. The effect is usually small and we are justified in 
assuming h„ — and in dropping the generator from the equivalent circuit. 

Values of 1/A„, approximate 100,000 SI and as such are negligible in effect 
when in parallel with load R values of 1000 to 10,000 CI. The product li„,R 
then ranges from 0.01 to 0.1 and h a ,R can be dropped in comparison to unity 
in the denominator of Eq. 4.17. 

By making h„ = and h at = 0, we obtain the simplified equivalent 
circuit for the C-E amplifier in Fig. 4. 12(a). The equation for the voltage gain 
is simplified from Eq. 4. 17 to 

" ~ V, ~ h„ 

The two assumptions made above may seem arbitrary; however, we 
should not expect close agreement between computed values and laboratory 
practice because of the considerable variation in the transistor parameters 
between units of the same type. Values given by the manufacturer are average 
parameter values. For example, while l\ f , may have an average value of 100, 
the actual values of this parameter may range from 50 to 200 in transistors of 
the same type. 

The simplified equivalent circuit is amply justified by the saving in work 
and in increased understanding of transistor circuit operation. 



(4.18) 



4.10 The Transconductance g n 



Equation 4. 1 8 for the C-E amplifier voltage gain employs the ratio h f ,/h lr . 
Reference to the /i-parameter definitions shows us that 



(4.19) 



The quantity I c IV b , is the reciprocal of resistance or a conductance with units 



78 



Junction Transistors as Amplifiers 



The Common-Emitter Amplifier 



79 



in mhos. It represents the output current obtained per unit of input voltage. 
Because it relates a current in the output circuit to a voltage in the input 
circuit, the quantity is a transfer conductance, called a iransconductance and 
given the symbol g m , where 



and from this can obtain the voltage gain as 



om is r 

K »» "I, 



(4.20) 



With g m — 0.01 mho — 10,000 //mhos, Eq. 4.20 indicates that we should have 
an output current of 10 mA ac for each ac volt input. 



!u. i 


J c *J±. 




+ 






+ 


v, '',>: 


■ c 




\ V > 






E 





!±+. B C Jl 




+ 




+ 


'•; h„i 


i 6 « 

8m V bc 


I k, 






E 





(3) 



(b) 



Figure 4.12 (a) Simplified C-E equivalent circuit (b) same as (a) with i! m Y tr as 
the source. 

In the equivalent circuit in Fig. 4.12(a), we see that the generator current 
and the collector current are the same and 

T, - h f Jt 

Using Eq. 4.20, we have 

h„h - g m V>, (4-21) 

We use g m V b , as the value of the current source in a revised C-E equivalent 
circuit in Fig. 4.12(b). The two circuits of the figure are completely inter- 
changeable, with one having its current source controlled by input current and 
the other by the input voltage. 



4.11 The Common-Emitter Amplifier 

We can now use the simplified g m model of the C-E amplifier in Fig. 4.12(b), 
to evaluate the performance of that amplifier circuit. 

Voltage Gain 

We first express the output voltage as 

V. = -I t R - -g m RV» 



A,. = jf = -g m R 



(4.22) 



Current Gain 






(4.23) 



The current in the load R is 

h "gmVu" h t>h 
by Eq. 4.21 and so 

Au — -j- = h f . 
This is the current gain of the transistor itself. 
Input Resistance 

By observation of the circuit, we have 

D 'I Y_bt L 

/ft 'ft 

for the resistance facing the input signal source. 
Output Resistance 



The resistance measured at the output 2,2 port is found by reducing the 
signal source to zero. With V, = V h , = 0, the collector current source is also 
zero, or open. From the 2,2 port we see an open circuit but in practice we 
know that !///„, is present there as a large resistance. Therefore 



(4.24) 



R " ^h~ 

"or 



(4.25) 



Power Gain 



The power gain of the circuit is defined as 

power gain = P.G. = M„/4„| 



(4.26) 
(4.27) 



This result is usually converted to decibels. The transconductance g m may be 
considered as a transistor figure of merit and its effect can be seen in the sev- 
eral gain expressions. 



80 



Junction Transistors as Amplifiers 



4. 12 Performance of a C-E Amplifier 

In a C-E amplifier we shall employ a pnp transistor with the following parame- 
ters: 

h it = 900 Q h f , = 25 

h„ = 2.1 x 10" ^ h ot = 16 x 10" 6 mho 3 




(a) (b) 

Figure 4.13 (a) Common-emitter amplifier: (b) with the simplified /i-parametcr 
equivalent circuit. 

The load R will be chosen as 2000 £2. 
With the circuit in Fig. 4.13, we have 

*» = fc = 9^-°- 0278mho 
The terminal resistances are 

R„ = h„ = 900 Q 
R . = -l- = 62,500 n 

The gains can be calculated as 

A,. = -g m R = -0.0278 X 2000 = -55.5 
A„ = h t . = 25 
P.G. = 25 X 55.5= 1387 
In decibels this becomes 

P.G. dB = 10 log 1387= 10 x 3.14 = 31.4 dB 

Had we used the exact expression for voltage gain, Eq. 4.17, we would 
have found the gain to be —53.8 compared with -55.5 above. The use of the 
approximate circuit and gain expression results in an error of only 3 per cent 
in the voltage gain. 

We can conclude that the C-E amplifier has moderate input and output 
resistances and good voltage and current gains. It is found to have the highest 






Conversion of the h Parameters 



81 



power gain of the three basic transistor amplifier circuits and is widely ap- 
plied for this reason. 

4. 13 Relation between A, and A r 

We can rearrange the C-E gain expression as 

(4.28) 



4.--fc»--fe*--4£ 



by reference to Eq. 4.20 and 4.23. Equation 4.28 is actually a general relation 
between voltage gain and current gain in amplifiers, which can be stated as 



. . load resistance ^ _ . R_ 

1 input resistance 'R, 



(4.29) 



This result may often simplify the calculation of the several circuit gain 
figures. 

4. 14 Conversion of the h Parameters 

The form of the /i-parameter equivalent circuit is the same for any two-port 
circuit, and therefore is the same for the C-E, C-B and C-C amplifiers. The 
parameter values, however, will vary with the choice of common element and 
Table 4.1 shows these relationships. 

Since the common-emitter parameters are most readily available, our 
circuit analyses will be carried out with those parameters. 

TABLE 4.1 C-B and C-C Parameters as Functions of C-E 
Parameters 



_ h„ 



*» - rrhj.-nj. 



n,b 



. hijlp, ll,, llf r 



I I h 



!• 



hie = hi, 

hoc = h , 

kf.- -(\+h„)s;-h f . 

h, c ~\ 



C-E and C-C Parameters as Functions of C-B Pa- 
rameters 



h,.= 



1 - a 



h,c = 



— Sfl 



l^a 



/. . . 'i/j/iot — h,t& 
"" 1-a 



L _ llcb 

h, c ~\ 



82 Junction Transistors bs Amplifiers 

4.15 The Common-Base Transistor 

Amplifier 

For ihe common-base (C-B) amplifier in Fig. 4.14(a), the input signal V, is 
supplied between emitter and base. The input current is /, and is much larger 
than the input current for the C-E circuit. This implies that we have a lower 
resistance transistor input circuit. We use the dashed box shown in Fig. 
4. 1 4(b) as the /;-parameter equivalent circuit for the transistor, with common- 
base h parameters. 




(O 

Figure 4.14 (a) Common-base amplifier; (b) the /i-parameler equivalent circuit; 
(c) equivalent g m circuit. 

Use of a conversion factor from Table 4. 1 allows us to write for the col- 
lector current 



From the C-B circuit in Fig. 4.14(b), we have 

V. b - h lb l. 



(4.30) 






The Common-Base Transistor Amplifier 

and using the conversion factor for h, b 



83 



We can determine the transconductance by taking the ratio of Eq. 4.30 to 
4.31, to give us 



'ft 



v ,b 



h„ 



I. 






(4.32) 



1 + V 

The transconductance for the C-B circuit has the same magnitude as for the 
C-E circuit; the negative sign arises because we are using V, b for the input 
voltage instead of V bl . 

Using -g m and h lb = hj{\ + h ft ) we draw the g m model for the C-B 
amplifier shown in Fig. 4.14(c). The performance of the C-B circuit can 
then be calculated, using the common-emitter parameters. 

Voltage Gain 

We have from the circuit 

h = -g m y, b (4-33) 

Because of the polarity assigned to the output voltage, we write it as 

v. - -/«* - g m RV,> 

Since the input to the transistor is also the signal input, or V , b — V„ we have 
for the voltage gain 



A* - fy = g m R 



(4.34) 



Current Gain 



The load current is given by Eq. 4.33 and if we use Eq. 4.31 for V rb , we can 
write l c as 

/ - &"> h <' l 

'°~ I M/,' 

The current gain can be obtained from this relation, as 



A _ h _ gJ'l. 

*» ~ % ~ • I hf. 



(4.35) 



We often find that h ft » 1 and the unity term in the denominator can 
be neglected, giving 



*,.= 



hftlhit 



= -1 



(4.36) 



as a limiting value. 



84 

Input Resistance 

The input resistance at the 1,1 port is 

/? _ hit. 



Junction Transistors as Amplifiers 



V An 



rnr B ?-^ (Q) 



(4.37) 



This is a low resistance, as predicted. 

Output Resistance 

We make V, = and effectively this reduces the current source to zero or 
an open circuit. The output resistance measured at 2,2 appears infinite but in 
practice it represents l//i o6 = (I |- h fc )/h oe , which is usually over I MQ. 

Power Gain 



This is obtained From A„ b and A, b as 



P.G. -\A,. b A lb \-g m R 



(4.38) 



The current gain approximates unity, with 180° of phase shift. The voltage 
gain is equal to that of the C-E amplifier in magnitude. The major reason for 
using this circuit is to obtain maximum power transfer through impedance 
matching. This is accomplished by matching the low-resistance input circuit 
to a low-resistance signal source and the high-resistance output circuit to a 
high-resistance load. Since the current from input to output is essentially 
equal, the circuit behaves in a manner similar to a water pump, forcing water 
from a low-pressure system to a high-pressure system. 



4. 16 Performance of the C-B Amplifier 

Using the same transistor as in the C-E amplifier example, we find the perfor- 
mance of the C-B circuit with a load of 2000 Q. 



h„ = 900 Q 
h„ 3 negligible 



h„ = 25 

h„, = 16 x 10" 6 mho 
= negligible 

We calculate g m = /,„//,„ = ffo ^ 0.0278 mho. 
The voltage gain is 

A tb = g m R = 0.0278 X 2000 = 55.5 
The current gain is 

A - *A. 0-0278 x 900 
*'"- TFry, 26" 



= - 0.96 






The Common-Co/lector Amplifier 

The power gain is 

P.G. = \A, b A lb \ = 55.5 x 0.96 = 53.4 
P.G. dB - 10 log 53.4 = 10 x 1.73 = 17.3 dB 

The input resistance is 

R » ~ ""■ T+T7, ~ "26 ~ J " 

and the output resistance is 

1 1 + h rt _ 26 _ , fi2 mq 



85 



h ol ~ 16 x 10- 
This is very large, as assumed. 

4.77 The Common-Collector Amplifier 

The common-collector (C-C) amplifier in Fig. 4. 1 5(a) is also called an emit- 
ter follower. This name is justified when we write the voltage equation around 
the input loop in Fig. 4.15(a) as 

V, - V b , - V„ = (4.39) 

Making the input signal V bt small, we have 

V,= V (4.40) 

The output voltage is approximately equal to, or follows, the input voltage. 
The voltage gain is, of course, near unity. 

The transistor is in the normal B, C, E connection and between those 
terminals we simply substitute the g m model of the transistor, as in Fig. 
4.15(b). The input resistance to the emitter is h„ and the current source is 

Km 'bf 

From the load circuit 

V. = (/* + hW (4-41) 




(a) (b) 

Figure 4.15 (a) Common-collecior amplifier; (b) g m equivalent circuit. 



86 



Junction Transistors as Amplifiers 



Performance of the C-C Amplifier 



87 






By definition in Sec. 4.10, we have 

I< = g«,V bt - h f ,l b (4.42) 

Using Eq. 4.42 in 4.41, we have 

K - (/» + h ft I b )R - (1 | h fc )RI b (4.43) 

Using Eq. 4.43 and V b , =- h,,l b in the input loop equation, Eq. 4.39, we have 

K <=[»,. + (1 I h rt )R)I h 
Input Resistance 

The input resistance at the 1,1 port of the amplifier is R le => V,\I b , which 
can be evaluated from the result above as 



•b 
Since h u > 1 is a usual condition, we then have 

R lc ^h„ + h,.R 
which is a large resistance, much greater than h lt . 



(4.44) 
(4.45) 



-o2 



-R„ 



-o2 



Figure 4.16 For the output resistance of the C-C amplifier. 
Output Resistance 

To obtain the output resistance at 2,2 we short-circuit the signal source 
V„ which reduces the control generator to zero or an open circuit. The circuit 
at the output port then becomes that shown in Fig. 4.16, with //„ in parallel 
with R, so 

""--KTR «■*> 

which is a low resistance, less than h„. 
Current Gain 

The output current in the load of the C-C amplifier is /, and we know that 

/. - -(/ b + I.) 






from the KirchhofT current law. But I, = //,,/» from Eq. 4.42 and the current 
gain can be obtained as 



•b >b 



(4.47) 



Voltage Gain 



as 



The voltage gain can be derived by use of the principle of Sec. 4.13, written 

"re — "te p 

hr.R 



*■•(' ' e«) 

using A, c from Eq. 4.47 and R le from Eq. 4.45. The expression above then 
becomes 

■ _ g m R 



A.- 



i i gji 



(4.48) 



which has a value less than but near unity. 

One use of the emitter follower is as a unity voltage gain amplifier to 
isolate one circuit from another; this might be the case if we supply a tele- 
phone line with the output of the amplifier. The C-C circuit is also widely 
used because of its high value of input resistance, which is much larger than 
can be obtained from a C-E amplifier. The C-C amplifier output also provides 
a match to a low-resistance load since R need be only 1000Q or so. This 
impedance transformation property, from a high input resistance to a low 
output resistance, is the inverse of the action of the C-B circuit. 

4.18 Performance of the C-C Amplifier 

Using the same transistor employed for the C-E amplifier, and with R = 
2000 Q, typical C-C performance can be calculated. The value of g m = 
0.0278 mho, as before. 



h„ = 900 n 


h,. = 25 


h„ ~ negligible 


/;„, = 16 X 10- 6 mho 


.*» 


= negligible 


The voltage gain is 




A - gn,R 
" 1 + gjt 


0.0278 x 2000 
1 + 55.5 


= 0.984 





and this approaches unity and has no phase shift. 



88 



Junction Transistors as Amplifiers 



The current gain is — h t , ■= —25, showing a 180° phase shift. 
The power gain is 

P.G. - | A VC A, C \ = 0.984 X 25 = 24.6 
P.G. dB = 10 log 24.6 = 10 x 1.39 
= 13.9 dB 

This is a low value of power gain. 

The input resistance is given by use of Eq. 4.44 as 

K = K + (1 | h f ,)R 

= 900 + 26 x 2000 = 52,900 £2 

Using the approximate relation in Eq. 4.45, we have 

R lc £ h„ + hf.R = 50,900 £2 

and the difference is not significant. 
The output resistance is 

h„R 900 x 2000 



*«- 



= 621 n 



h le + R 2900 

which is quite low, as predicted. This design would be well suited to operate 
from a high-resistance microphone into a telephone line. 



4.19 Comparison of Amplifier 
Performance 



Table 4.2 is introduced to summarize the predicted performances of the three 
basic amplifiers and to allow comparison of their respective capabilities. 



TABLE 4.2 Comparison of Basic Transistor Amplifiers 



C-E 



C-B 



C-C 



Input resistance, R/ 


h„ 


h„ 




1 + h„ 


Output resistance, R 
Current gain. A, 


I 




Voltage gain, A. 


gmR 


gmR 


Power gain 


ll/,gmR 


gmR 


Power gain as a ratio to that 
of the C-E amplifier 


1 


1 



hi. + h ft R 

h„R 
hi. + R 

at l 

hf.gmR 

1 +g m R 

I 
1 +g„R 



Table 4.3 adds the data from the several examples, again for comparison 
purposes. The data show that the C-E circuit has nominal input and output 



Transistor Manufacturing Techniques 



89 



resistances and the highest power gain. For those reasons it is the most 
generally used amplifier circuit. The C-B circuit finds application as an imped- 
ance matcher from low to high values, while the C-C circuit isolates the input 
from the output and matches high to low impedances. 



TABLE 4.3 Typical Transistor Amplifier Performance 



Transistor 


: h lt 


= 90on 


h/. = 25 






h„ 


= negligible 


h„ = 16 x 10-« 


mho 




Km 


= 0.0278 mho 


2s negligible 








C-E 


C-B 


C-C 


Hi 




900 n 


35 a 


52,900 n 


Ro 




62,500 n 


1.62 Mil 


621 fi 


A, 




25 


-0.96 


-25 


A,. 




-55.5 


55.5 


0.984 


P.G. 




1387 


55.5 


24.6 


P.G. dB 




31.4 dB 


17.3 dB 


13.9 dB 



4.20 Transistor Manufacturing 

Techniques 

The manufacture of transistors employs the same basic purification and 
crystal growth processes already described for diodes. Subsequent processing 
of the semiconductor material leads to grown junction transistors, alloyed 
junction transistors, and mesa and planar transistors. 

Grown junction transistors arc produced by crystal growth from a doped 
bath of molten semiconductor, as diagrammed in Fig. 4.17(a). Change of the 
predominant impurity from p to n and back to p is made at precise moments 
to yield layers of the desired conduction properties. 

Wafers are then cut from the crystal to include sections of the junction 
regions. Critical resistance requirements for the several regions are hard to 
meet and the method is not widely used. 

Alloyed junction transistors are made by fusing a pellet of the desired 
impurity element onto both sides of a base chip, as in Fig. 4.17 (b). With an 
n-silicon base,/? impurity pellets of indium will alloy with the silicon to form 
injunctions at the interfaces, as for the alloyed diode. Precise control of time 
and temperature of the fusion process yields transistors in which the critical 
base width can be very thin. The remainder of the impurity pellets serves as 
the emitter and collector contacts. 

The process of diffusion is widely used in the manufacture of mesa and 
planar transistors. A substrate wafer of desired resistivity is coated with a 






90 



Junction Transistors as Amplifiers 



Cryslal 
Pulling 
Machine 



3? 



P 



melt 



(a) 





Epitaxial 
iy p Layer 



E B C 



Oxide Layer 
£ 




(c) 



(d) 



Figure 4.17 (a) Grown pnp transistor material ; (b) alloyed pnp junction transis- 
tor; (c) diffused mesa transistor; (d) planar transistor. 

layer of a doping element, by deposition from a gaseous atmosphere contain- 
ing silicon and the desired doping material. Exposure to high temperature for 
a limited time follows while the doping atoms migrate to the desired depth in 
the wafer. If an n-silicon wafer is used with boron as the diffusing impurity, 
the surface of the wafer will become/) material to the desired depth. A second 
diffusion with an n impurity changes some of the p material back to n for an 
emitter layer. The result is illustrated in Fig. 4.18. The depths of diffusion are 
a few thousandths of a millimeter. 

The diffusion process can be confined to very small areas by use of pho- 
tographically applied chemical surface masks. The surface of the silicon is 
coated with a very thin layer of silicon dioxide, followed by a layer of photo- 
resist lacquer. The mask is placed over the wafer in accurate register and 
identical areas for all the transistors are exposed to ultraviolet light. Where 
exposed to light the photoresist is chemically changed and made inactive. The 
unexposed areas are chemically etched away along with the underlying silicon 
dioxide, leaving openings to areas of the silicon wafer. The silicon of these 






Transistor Manufacturing Techniques 



91 



all 

- « H 

E C o 



I0'< 


L 


.-" 


I0'° 

p' 

10* 




Collector Impurity 
Density 

7..... 


c 



0.002 0.004 0.006 

Distance from Surface (mm) 
Figure 4.18 Diffusion process to form p base and n emitter. 

electrode areas is available for further impurity diffusion. The process is 
repeated with different masks, resulting in superimposed electrode areas. The 
masks are similar to negatives, produced as much-reduced photographs of 
large-scale drawings. The masked areas are very sm'all and hundreds of tran- 
sistors can be produced on one silicon wafer cut from a 2- or 3-cm diameter 
crystal. 

In the process of epitaxial growth (epi, upon; taxi, arrange), a doped 
crystal region is grown upon the wafer with the same atom arrangement and 
crystal structure. Growth usually occurs from a vapor containing silicon and 
a doping material at an elevated temperature. The impurity concentration can 
differ from that of the substrate, giving lower or higher resistivity in a thin 
layer. 

Use of a high-resistivity collector wafer raises the breakdown voltage but 
also increases the saturation resistance and the collector voltage drop at high 
current. By epitaxial growth, it is possible to place a high-resistivity but thin 
collector layer on a low-resistivity wafer, giving both high breakdown voltage 
due to the high-resistivity material at the junction and a low saturation resis- 
tance due to the bulk of the collector material. The use of an epitaxial layer is 
shown in the mesa transistor in Fig. 4.17(c). 

The mesa transistor may use an epitaxial layer on a p wafer, followed by an 
fl diffusion to form the base region. A second diffusion is applied to a small 
area through an unmasked opening in the silicon dioxide and a small p 
region formed for the emitter. Excess n material is then etched away to reduce 
the size and electrical capacitance of the collector-base junction. The name 
mesa is derived from the similarity of the completed transistor to the desert 
mesas of the western United States. 

The planar transistor in Fig. 4.17(d) is produced by diffusion of an n 
impurity to the desired depth in the p chip through openings in a silicon 
dioxide layer, masked as described above. The process is repeated and a p 
impurity is diffused to form the emitter region in the previously diffused base. 



92 



Junction Transistors as Amplifiers 



Silicon dioxide is added to reduce surface leakage and aluminum is evaporated 
through openings in the dioxide layer to form the necessary clement connec- 
tions. 

4.27 Comments 

The junction transistor represents two back-to-back diodes, with the signal 
voltage controlling the current through the forward-biased input junction. 
This current is extracted through the reverse-biased output junction as a cur- 
rent source. 

In Chapter 3 we developed algebraic methods of active circuit analysis 
and a general //-parameter equivalent circuit model. This circuit, being equiva- 
lent to any active linear circuit, is used as an ac equivalent for the transistor 
by assignment of/; parameters measured at the ports of the actual transistor. 
The transistor becomes a set of circuit elements and a source, and for ac 
signal analysis we have no concern for holes, electrons, junctions, and 
biases. Since the output of a transistor is a varying current and we require a 
voltage for input to a cascaded transistor, we use a load resistor to provide 
conversion between output current and output voltage. 

There are three basic amplifier forms, the C-E, C-B, and C-C, for which 
we employ the same equivalent circuit model. This is the g m model for the 
active source, in which the internal current is made dependent on the input 
signal voltage. This g m model will be applied for the field-effect transistor 
and vacuum tube as well. 

REVIEW QUESTIONS 

4.1 Draw the circuit symbol for a pup transistor; show and label the three assumed 
currents and the three voltages. 

4.2 Repeat Question 4.1 for an npn transistor. 

4.3 What is meant by I c , I B , V b „ I c , I b , V BB , I E , V cc 1 

4.4 Explain the meaning of npn and pnp when applied to transistors. 

4.5 What is the collector polarity with respect to the emitter in a pnp transistor? 

4.6 What bias polarities should be applied to the base and collector of an npn 
transistor, with the emitter as reference? 

4.7 What kind of charges transit the base in an npn transistor? 

4.8 Why are the charges injected into the base able to reach the collector terminal 
in a properly biased transistor? 

4.9 Why is the impurity concentration made much greater in the emitter than in 
the base? 

4.10 Why does I CBO vary with temperature? 

4.11 What should be the allowable maximum temperature of a silicon transistor to 
keep I CB o small? 



Review Questions 



93 



4.12 The transistor is sometimes said to be a current-operated device; use the diode 
law and explain that it is also a voltage-operated device. 

4.13 What currents and voltages are related by the output characteristics of a tran- 
sistor with common emitter? 

4.14 On the input curve of Fig. 4.5, choose a Q point and identify it. What should 
V BB be to obtain that Q point? What would I B then be? 

4.15 What is the cause of / ca0 ? 

4.16 For a C-E transistor, define h rE in words. 

4.17 Define the current gain a in words. 

4.18 What is the approximate relationship between h PE and h f , at usual operating 
temperatures? 

4.19 Define the saturation resistance for a C-E transistor. 

4.20 Explain how the h parameters arc determined from the characteristic curves. 

4.21 Why can h FB never be greater than unity? 

4.22 If a is given, how do you determine /i/ f ? - 

4.23 If hfE = 60 and l c — 15 mA, what is the base current? What is the emitter 
current ? 

4.24 What physical measurement of the transistor should be reduced to increase 



4.25 State one reason for choice of a C-B amplifier. 

4.26 State one reason for choice of a C-C amplifier. 

4.27 What is the major advantage of the C-E amplifier over the other forms? 

4.28 Define A m A,, and P.G. What does the negative sign on a voltage gain figure 
mean? 

4.29 Why is the input resistance of a C-B amplifier so much smaller than that of 
a C-E amplifier? 

4.30 Compare the input resistance of a C-C amplifier to that of a C-B amplifier. 

4.31 Why is the C-C circuit called an emitter follower? 

4.32 A C-B circuit is sometimes called a collector follower; explain. 

4.33 What circuit element value should be changed to make A vc approach unity in 
a C-C circuit? 

4.34 How should a transistor be chosen to accomplish the purpose of Question 
4.33? 

4.35 What assumption has been made to reduce the C-E amplifier voltage gain to 
-gmRI 

4.36 Describe the fabrication of an alloy junction transistor. 

4.37 What is meant by the epitaxy process? 

4.38 What is a mesa transistor? 



94 Junction Transistors as Amplifiers 



PROBLEMS 

4.1 For ihe transistor in Fig. 4.8, find /« Tor a Q point at / c = 15 mA, with Vcc = 
12 V and a load of 500 ft. 

4.2 Find l B for the transistor in Fig. 4.8, with V cc = 30 V and R 1200 ft, V CB 
= 15 V. 

4.3 For the transistor in Fig. 4.8, the dc load line is drawn between V cc = 20 V, 
l c = and / c - 30 mA, V CE = 0. 

(a) What value of R is being used in the circuit? 

(b) What is l c if the Q point is selected at l B - -- 100 /iAl 

4.4 Determine h FE for V CE = 20 V, 1 B - 250 #A for the transistor in Fig. 4.8. At 
the same Q point, what is the value of a? 

4.5 A silicon transistor has A; c m 1.80 mA for A/ £ = 1.89 mA. What change in 
/'« will produce an equivalent change in i c f 

4.6 The transistor described by Fig. 4.4 and 4.5 is biased by V BB ■ 0.9 V at 0°C. 
Determine l c and V CE for R 6000 ft, V cc = 30 V. Hint: Find the Q point 
from Fig. 4.5. 

4.7 A transistor in a C-B circuit has I„ 105 /;A, / c = 2.05 mA. Determine h FE ; 
also find 7 £ . 

4.8 For the transistor of Problem 4.7, we note that when i a changes i 27 //A, fc 
changes +0.65 mA. Find a and lt ft . 

4.9 The >/ parameters for a transistor are measured as 

/;„ - 800 ft g m = 0.035 mho 

h„ - negligible //„, = 9 X 10" 6 mho 

With R 5000 ft and neglecting li or , find A„ A,, R,, and the power gain in 
decibels in a C-E circuit. 

4.10 Determine R lr , A„, and A u for a load of 1000 ft in a C-E circuit with a tran- 
sistor having 

/;„ = 1500fi A/, =40 

/;„ = negligible h ot = 5 x 10 -6 mho 

What error is created in A,., by neglect of //„,? 

4.11 Find the power gain in decibels when the transistor of Problem 4.10 is used in 
the C-B circuit with R = 5000 Q. 

4.12 Repeat Problem 4.10 for the C-C circuit. Also find R oc . 

4.13 With the transistor of Problem 4.9. determine A rb , A lb , R lb , and P.G. dB in the 
C-B circuit with R - 1500 £2. 

4.14 In a C-C amplifier we use R - 1500 Q with a transistor having 

/i„ = 1800Q gm = 0.018 mho 

h, e = negligible h , = negligible 

Find A vc , A, c , R, e , R oc , and P.G. dB . If h Bt = 9 x 10 -6 mho, find the per cent 
error in neglecting it in computing A lc . 



Problems 



95 



4.15 A transistor having h, t = 65, h„ = 850 Q. h„ negligible, and /;_, neg- 
ligible delivers an output of 6 V across a 1500-£2 load resistance in a C-E 
amplifier. 

(a) What is the ac input current? 

(b) What input voltage is needed? 

4.16 A transistor having h„ = 1200 ft, h f , = 32, h„, = negligible, and h„ = neg- 
ligible is used for Q { and Q 2 in the circuit in Fig. 4.19(a). Determine the current 
gain for these transistors in cascade with R = 2000 fi. Find the overall A r . 
Hint: The load of Q, is the input resistance of Q z . 





(a) 



(b) 



Figure 4.19 



4.17 Wc have a signal source with an internal resistance of 150ft connected to 
a C-B amplifier. What value of input resistance should the transistor have for 
power matching? With an input signal of V,„ - 0.1 V, what value of# m should 
the transistor have to give an output of 12 V across a load of 2500 ft ? 

4.18 A signal source of 100,000 ft internal resistance is to be power matched by the 
input circuit of a C-C amplifier. IT //„ - 800 ft and g m - 0.025 mho, what 
value of R should be used? What is the resulting /!,.? P.G. dB ? 

4.19 Draw thcg m equivalent circuit for Fig. 4.19(b) when using a transistor having 
h„ = 1250 ft, h f , = 35, h„ = negligible, and /;„, - negligible. What output 
resistance will be measured at the 2,2 port? What is the voltage gain? 



Choice of the Quiescent Point 



97 



5 

DC Bias for the 
Transistor 



Transistor amplifiers must be operated with steady bias voltages and currents 
to provide the desired operating conditions for the emitter-base and collector- 
base junctions. The bias sources also supply the circuit energy. 

The transistor converts only a part of the dc input energy to ac signal 
output; the remainder is dissipated as heat and must be removed. This ther- 
mal loss and the resultant temperature rise place a limit on transistor opera- 
tion. 

The maintenance of the steady currents and voltages through circuit 
design will be discussed in this chapter. Biasing circuits are added to the 
amplifiers of Chapter 4, hopefully without reducing amplifier performance. 
We find, however, that obtaining stability for the bias collector current often 
causes a loss in performance. 

5. 1 Choice of the Quiescent Point 

On the output characteristics, the usable region for transistor operation 
is defined by the maximum ratings of the transistor. In Fig. 5.1 this usable 
region is bounded by the following: 

1. The horizontal line at maximum allowable collector current. 

2. The saturation line for the transistor near v CB = 0, where the base 
current loses control. 

96 






3. The cutoff line at i„ = near the abscissa. 

4. The maximum allowable value of collector-emitter voltage to prevent 
avalanching in the reverse-biased junction. 

5. The hyperbolic curve set by the maximum safe transistor power dis- 
sipation. 

The maximum dissipation curve results from the transistor power rating 
for maximum allowable temperature, and the curve is drawn for 

In Fig. 5.1 this maximum dissipation curve is drawn for P d -^ 200 mW. 
For large power output the Q point will be located close to but below this 
maximum rating curve. 




Figure S.I Choice of a Q point for a transistor; R = 3750 CI. 

For linear operation, the Q point should be placed near the central por- 
tion of the region where the curves are most linear and uniformly spaced. 
For small-signal linear operation, there is no unique location for the Q 
point but signal swings along the load line should remain within the linear 
region. Any Q point in a region having constant h„ h f , and lt values will 
give equal performance for a given load. 









98 



DC Bias tor the Transistor 



Variation ol the Q Point 



99 







'c 


/A\ 


h 
















1 \ fN 




i X / \ v* 


,T\ 


*" \ xX 


YT" 


Y"—-. 


— x ^ 



"a 



(h) 



Figure 5.2 (a) Positive peak clipping; (b) negative peak clipping, 

A sine input signal applied to a Q point at A, Fig. 5.2(a), can cause output 
distortion at the positive peak, due to transistor saturation. A sine input 
signal applied at the Q point at B, Fig. 5.2(b), can drive the transistor into 
cutoff on the negative swing and distortion of output waveform results. 
At a central location at Q on the output characteristics, saturation or cutoff 
distortion is not likely to result for small-signal amplitudes. For this purpose 
we define a small signal as one whose amplitude is small when compared to 
the base-emitter bias voltage or current at the Q point. For V BE = 2 V, we 






might consider an input sine signal of +0.2-V peak, or less, to be a small 
signal. 

5.2 Variation of the Q Point 

The current ratio h rE varies with temperature over a considerable range for 
a given transistor, as shown in Fig. 5.3(a). In addition, h FE is not closely 
controlled in manufacture and a typical rating might be h FE = 60 but with 
a range from 30 to 110. Since 

lc - h FE I B 
it can be seen that the Q point at l c ^ 4.6 mA, for I B = 80 fiA, could not 
be maintained for different transistors. For a fixed-bias current of 80 //A, 
the g-point current would range from 2.4 to 8.8 mA for various transistors 
of the same type. The latter value would lead to a saturation situation. 

In small-signal amplifiers there will usually be sufficient load resistance 
in the collector circuit to hold l c within safe limits. A change in h FE will 
cause a change in l c and l B , however. This may result in distortion through 
movement of the Q point toward cutoff on the input curve, Fig. 5.3(b). 
Cutoff distortion may also be created by further movement of the base- 
emitter bias point due to temperature changes. Fixed-base bias voltage, 
V BB = 0.84 V, will allow the base bias current to change from 40 to 240 //A 
as the temperature changes from to 50°C. 

The design of the base circuits should provide some stabilization of the 
g-point currents against changes in h FE and v BE due to manufacturing 
variations in transistors or to changes in temperature. 

The transistor reverse current J CBO is a sensitive temperature function, 
nearly doubling for each 10°C (18°F) rise in transistor temperature. In power 
amplifiers this can cause damaging increases in l c because of the small 
amount of resistance in the collector circuit. Methods of protection of the 
transistor against such damaging currents will be discussed in Chapter 12. 

To convert readily from one transistor current to another, we show their 
relationships in Table 5.1. 

TABLE 5.1 Transistor Current Relations 



From 
Ib 
lc 






To l B 

Multiply by 

I 

I 

*Te 

\ 

1 I hn 



hFE 

1 

I'FE ~, I 

1 hy E 



I + h fE ss h rE 

I + h, B ~ 1 
hhE 

1 



100 



220 
200 
1 80 
1 60 



jS I40 

1 20 

1 00 

80 

60 



DC Bias lor the Transistor 



20° 0° 20° 40° 

Temperature (°C) 
(a) 

300 



200 

























































"fe/ 









































































































60° 



< 



100 























1 










50°C 


II i 
]0°c/ 




































I 







2 


4 


6 0. 


A i 






(b) 

Figure 5.3 (a) Variation of7/f £ for a silicon transistor; (b) variation of v BE with 
temperature. 

5.3 Fixed Transistor Bias 

The circuit in Fig. 5.4(a) is perhaps the simplest form for base-current bias 
but it is the least satisfactory in compensation for changes in h FE . It can, 
however, effectively reduce bias current changes due to temperature varia- 



Fixed Transistor Bias 



101 



tion of V BE . Writing a voltage equation around the input loop of Fig. 5.4(b), 
we have 

Vcc - RbIb V be - 
and we solve for the base current as 



r _ ' cc 

• H 



V rr ~ V„ 



R„ 



But because V cc » V BE , we have 



J ~ 'cc 

Ib = ic b 



(5.2) 



(5.3) 



This shows that changes in V BE due to temperature have little effect on the 
bias current I B if V cc is large. The choice of R B then determines the bias 
current for the base. 




A 




-=r Vcc 



^ 



(a) (b) 

Figure 5.4 Fixed-bias circuit analysis. 



(5.4) 



The following equation may be written for the output loop: 

V C c - Rlc -Vc* = 
But by definition of h FE 

'C = "FE' B 

and so transposing and substituting in Eq. 5.4 we have 

Vce — V C c ~ R?c 

= V cc — h. . Rl 

With V cc and R fixed and l B a constant bias current, then Eq. 5.5 shows that 

a change in h FE must change V CE and this causes a shift in the Q point. 

It would be necessary to adjust the bias current by changing the value 

of R B in each amplifier completed in a production run. Readjustment of R B 



(5.5) 



102 



DC Bias for the Transistor 



would be necessary lo compensate for variations of h lE every time a new 
transistor were placed in the circuit. 

It should be obvious that a better bias circuit is needed, one that will 
prevent shifts of the original Q point for changes in h FE and V BE . 



5.4 The Four-Resistor Bias Circuit 

The bias circuit in Fig. 5.5(a) uses four resistors and provides improved 
stability of the Q point. The input circuit resistors /?, and R x are used to give 
fixed bias, while the emitter resistor R E provides a bias voltage that varies 
with I c . This compensates somewhat for changes of h PE . 

Consider first the input circuit, which is redrawn in Fig. 5.5(b). We assume 
I B <S /, or that /, = I 2 . If I„ is in the microampere range and if /, and /, 
are adjusted to the fractional milliampere range, then our assumption is 
valid. A suitable selection is to make R t + R 2 = 60,000 to 100,000 fi. 

At the B, E terminals we can replace /?, and R 2 and the voltage V H with 
a voltage-source equivalent circuit. With R, and R 2 in series and l B negli- 
gible, the open-circuit voltage at B, E is given by the voltage-divider ratio 



V nr = V R = 



R* 



/?, + R 2 



(5.6) 



If we short-circuit the B, E terminals, the short-circuit current becomes 

(5.7) 
Then, by our rules for deriving the voltage-source equivalent circuit, 






R' = Re = -7* = 



__M> 



(5.8) 



VcdRx ~ Rx + *2 

This is the value for /?, and R 2 in parallel. The resultant voltage-source cir- 
cuit is drawn at the input in Fig. 5.5(c). 

Writing a voltage equation around the input as redrawn in Fig. 5.5(d), 
we have 

-V B + R B I B + V BE - R E I E - (5.9) 

We can replace I E with 

1b - "(/* + Ic) - -(I I h FE )l B 
and have 

-V B + R B l B I V BE -i (1 + h FE )l B = 
Solving for the base current, we find 



/„ = 



V.-Vi 



BE 



V*-Y* 



Rb-tO +h FE )Rt 



I 



, _,_ (! +h FE )R E 
+ ' Rb 



(5.10) 









The Four-Resistor Bias Circuit 



+ 0^a 




(a) 




lo- 



103 



'l||K 



lo |f 



h \ 



-o 



+ 

-=- y<x 



Rifv B -^. 



T 



(b) 




>'«,- + **iU 



(0 (d) 

Figure 5.5 (a) The four-resistor bias circuit; (b), (c), and (d) reduced circuits. 



For the fixed-bias circuit 

Ic-h FE l B (5.11) 

and for the four-resistor bias circuit, I c is h FE times the I B value in Eq. 5.10, 
leading to 



/r = 



y„-y» 



Rb 



, , (1 \-h FB )R B 
M Rb 



We can make I c less affected by temperature-induced changes of V BE 
by making V B larger. Since V BE approximates 0.5 V, we may make V„ — 3 V, 






104 



DC Bias for the Transistor 



as an example. We may also neglect unity with respect to h FE and then have 






l+h, 



<7T M } 



(5.12) 



With the fixed-bias circuit, I c was given by Eq. 5.11, and was directly 
affected by changes in h FE , with the bias current held constant. With Eq. 
5.12 for the four-resistor circuit, however, we have h FE appearing both in 
the numerator and the denominator. It may be seen that the sensitivity of 
I c to changes in h FE can be reduced if we choose R E IR B correctly. 

If we make R E = 0, however, then Eq. 5.12 reduces to 



/ — ^ B h 
'c — Tr" Ft: 
Kb 



(5.13) 



and I c will vary directly as h FE , which is the undesirable situation in the fixed- 
bias circuit. In fact, making R E = in the circuit in Fig. 5.5(c) reduces it 
to the fixed-bias circuit. 

If we make R E /R B ~ 1, then the li FE (R E /R B ) term in the denominator 
would be the influencing factor. The h FE terms in numerator and denominator 
would approximately cancel and l c would be independent of h FE , which is 



a o 









© 


1 






4 






9 














/ 














i 






pO.Oip 


i 




/ 


— 






. % 




/ 




y 


s 








/ 


y 


* 








/ 


/ 


/ 








7 


/ 


/ 








0.03 
0.05 


m 


// 












Vs 










n 1 — 


/ 


'^ 










U. 1 


J 










1.0 

— 1 — 



20 40 60 80 100 120 

Figure 5.6 Curves for increases in l c due to change in li FE . 






Design of a Fixed- Bias Circuit 



106 



the result desired. R E must be small in magnitude or we lose signal gain and 
dc power, however, and R B must be large with respect to h„ of the transistor 
with which it is in parallel or we lose gain again. Thus some compromise 
between ideal bias stability and gain performance must be made. 

The sensitivity of J c to variations in li FE , for various values of R E /R B , is 
plotted in Fig. 5.6. Values of R E /R B in the range of 0.05 to 0. 1 usually lead 
to satisfactory gain and reasonable stability against changes in h FE . With the 
R E /R B ratio equal to 0.1, the value of I c will increase 70 per cent when h FE 
changes from 20 to 100. Without the compensation provided by R E , the value 
of I c would increase as shown by the curve for R E /R B = 0, or 500 per cent 
for the same h FE variation. 

Thus we have the design conditions for the four-resistor bias circuit: 

K fl = 3V 

5^ = 0.05 to 0.I 
Kb 

We call R E /R B , or its equivalent in other circuits, the stabilizing ratio. This 
ratio serves as an index of stability, which increases as / c is made more stable 
by the circuit design. Figure 5.6 functions as a universal curve for predicting 
such stability as a function of the stabilizing ratio. 



5.5 Design of a Fixed- Bias Circuit 

The fixed-bias circuit in Fig. 5.4(a) will be designed, using the transistor with 
characteristics of Fig. 5.7, with h FE - lOO. A Q point at V CE = 10 V, l c = 
2 mA is chosen as a desirable location in the middle of the linear region of 
the curves. We usually assume that V cc = 2V CE and so we have V cc = 20 V. 
The load resistor can be found from the voltage equation around the 
output loop, Eq. 5.4; 

V cc ~ Rlc - Vce - 



R = cc 



Ic 

20- 10 



(5.14) 



"0.002" 



= 5000Q 



The remaining resistor is R B and the base current is 

Ic 



/„ = 



0.002 
100 



= 20 x 10"«A = 20^A 



This value could also be read from the g-point location at V CE = 10 V, 



106 



DC Bias for the Transistor 



Design of a Bias-Stabilized C-E Amplifier 



107 



60 
50 
40 



< 

a. 30 



20 
10 









' 
















f2) — 












i 






— ' 


/! 





< 
£ 4 



r 


.oad L 


JflJ 


60 ^ 




r ix- 


inc 




50 

r£ — 


* 










v 














r 


\ 








30 


' 




\ 






20 


rl 




\ 


s t 




10 


rl 






\ 








0.5 

"BE 
(a) 



1.0 







30 



10 20 

Has 

(b) 
Figure 5.7 Curves for the design example. 



I c = 2 mA in Fig. 5.7(b). Rewriting Eq. 5.2 we have 

p "cc ~ 'BE 

K B — J 



(5.15) 



For a base current of I B = 20 ftA at the Q point, we can go to the input 
curves of Fig. 5.7(a) and find that V BE at (I) must be 0.65 V. Then 



* H - 



20 - 0.65 



19.35 
20 x 10* 



20 x 10-« 
= 967,000 CI S 1 Mfi 

and our circuit design is completed. 

In the analysis of the circuit, an assumption was made that a large V cc 
would stabilize the circuit against changes of V BE with temperature. This can 
be shown by allowing V BE to be zero volts, in which case 

20 



* B - 



20 10 " 



= 1 MQ 



Then we can make V BE = 1 V and find that 



R„ = 



20- 1 
20 X 10"* 



- 0.95 Mfi 



The required values of R B at 1 Mf2, 0.967 MQ, or 0.95 MQ all lie within 
the tolerance zone of a 1-Mf2, rhlO per cent resistor so that changes due 









to extreme values of V BE cause less circuit change than might be expected 
to occur in normal production runs from the tolerance on ^? a . 
With 

Ic = h fE I B (5.16) 

and l B fixed by our circuit design at 20 /xA, however, the collector current 
I c will vary directly with changes in h fE . Since the latter parameter is sensitive 
to temperature and to variation between transistors of the same type, the 
fixed-bias circuit gives no stabilization of the Q point with respect to fl, B 
changes. 

5.6 Design of a Bias-Stabilized 
C-E Amplifier 

A bias-stabilized network for a C-E amplifier, using the transistor in Fig. 
5.7, will be designed. This transistor has h„ = 650 SI and h tE = 100. Pre- 
viously it was pointed out that there is no unique location for the Q point 
for a linear amplifier since with a small signal and constant h fe , h„, and h ot 
values over a region of the transistor curves we would have equal amplifier 
performance at many Q points in the central linear region. Arbitrarily we 
select a Q point at I c £■ 3.2 mA along the I B = 30 fiA line. A supply voltage 
less than the rated K C£(n ,„, should be chosen. A value of 20 V was selected. 
The values for the four resistors, one being the load resistor, are to be 
determined for the circuit in Fig. 5.8(a). There are more circuit unknowns 
than there are relations to solve and so we must call on engineering experi- 
ence to determine some of the design values. 

+ 20 



9Vr, 





(a) (b) 

Figure 5.8 (a) Four-resistor C-E amplifier; (b) output circuit and currents. 



108 



DC Bias lor the Transistor 



Through the collector circuit, isolated at Fig. 5.8(b), we write a voltage 
equation 

Vcc - Me ~ Vce - RbVb i lc) =- 
Vcc - V CE - (R + R E )J C + Rah 
But since I a <C / c , we can then simplify the above to 

Vcc ~ Vce S (/? i *j)fc (5.17) 

We now have one equation and three unknowns, V CB , R, and R B . Experience 
indicates that the voltage across the transistor at the Q point ought to be 
about one-half of the supply voltage, expressed mathematically as 



Then Eq. 5.17 becomes 



v — ■ cc 



V -f = WR + R B ) 



(5.18) 



We have now reduced the unknowns to two. 

For our particular amplifier, with I c — 3.2 mA, we can write 

10 - 0.0032(K I R E ) 
10 



R f R E - 



0.0032 



= 3100Q 



Since usual resistors are accurate to ± 10 per cent, we round off the numerical 
result. 

We can now draw the dc load line for R + R B on the transistor charac- 
teristics, from V cc = 20 V to i c = V cc j(R I R t ) = 6.5 mA, as intercepts. 
The Q point is placed at I„ = 30 fiA, V CE = 10 V as selected. 

The division of R + R E = 3100 Q. between collector and emitter circuits 
must now be determined. The emitter resistor was added to stabilize the 
steady collector current against variations in h FE . This purpose would dictate 
a large value for R E . The dc voltage across R E reduces the voltage across 
R, however, and the latter voltage limits the peak swing of the signal voltage 
across the load. Thus R E should not be large. As a rule, we make the emitter 
voltage V E about 10 to 20 per cent of the supply voltage and for this design 
we shall choose 

= 2V 



(5.19) 



We then have 



R E = L*^L* 



l E 

2 
0.0032 



lc 
^600fi 



(5.20) 






Design of a Bias- Stabilized C-E Amplifier 109 

Consequently, 

/? = 3100- 600 = 2500 

and the output circuit is complete. 
The base voltage V B is 

V B = V B + V BE (5.21) 

We go to the input curve, Fig. 5.7(a), and for a base current /„ — 30 //A, we 
find at (2) that V BE is 0.70 V. With 

V -~ R ^ (5 22) 

= 600 x 0.0032 = 1 .92 V s 2.0 V K ' 

Then, using Eq. 5.21, 

V„ - 2.0 -f 0.7 = 2.7 V 

We are able to write two design equations involving resistors R, and /J 2 : 






R,+Ri 
Rz ■ 



(5.23) 
(5.24) 



R:+R 2 ' cc 

We hope to make /, > 10/ a so that the current I B can be neglected in passing 
through /?,, as we assumed in the circuit analysis. Therefore, we choose 
/, S / 2 as 10 X 30 //A = 0.3 mA and from Eq. 5.23 we have 

«H Rz-^fz 
' i 

= 3-3rTo^ 67 ' ooon 

From Eq. 5.24 we can find R 2 as 

V» 



(5.25) 



R, = (R, i /?,)- 

' cc 

= 67,000^ _ 9000 Q 



(5.26) 



Then 



/?, = 67,000 - 9000 = 58,000 fi 

The design of the circuit is now complete and it is shown in Fig. 5.9. 

The bias resistor /? 2 is in parallel with h lt — 650 £2 of the transistor, and 
the shunting effect of/?, is negligible. 

That l c is stabilized can be determined. The value of R„ can be found from 
Eq. 5.8 as 

__RjRj_ 
R,+Rz 
58 x 10 3 x 9 x W 



R B - 



67 x 10 3 



= 7800 



110 



DC Bias tor the Transistor 



The stabilizing ratio R E fR B is 



S. R. - 4^ - ^ - 



600 



R, 



7800 



Reference to Fig. 5.6 for h rE => 100 shows that the collector current can 
change only by a factor of 1/1.6 if h fE falls from 100 to 20. Thus our basic 
design choices are shown to be satisfactory, using the stabilization criterion 
of R t: IR„. 



r cc 



lb 



\lr 



lo lr- 




H( °- 



•c\ 



V. 



-o2 



lo- 



Figure 5.9 Completed design for the four-resistor bias circuit. 

5. 7 Voltage Feedback Bias 

The circuit in Fig. 5.10 uses an emitter resistor for stabilizing l c and adds 
further control through variation of 1„. If l c increases, the RI' C voltage drop 
increases, lowering the voltage at the collector. Resistor R f supplies base 
current from the collector and as I c increases, the base current falls. This 
change in I„ tends to oppose the original change in I c . 

Around the input loop, shown in heavy lines in Fig. 5.10, we have the 
voltage equation 

V cc - Rl' c - R,I B -V BE - R E l' c = (5.27) 

Since l„ is small compared to l c , we can neglect the branching of l B at point 
A at the collector and say 

£=/ c (5-28) 

Rearranging Eq. 5.27, 

V CC -V BE = (R I R E )I C +Rfh 

We can drop V BE as small compared to V cc . Then, using I B = I c lh rE , we can 
write the equation in terms of l c as 



rcc- (* + *, + jj£)lc 



Voltage Feedback Bias 



111 



+ :ov 




Ol 



Figure 5.10 The voltage-feedback bias circuit. 



We can solve for I c and obtain 

/, Vcc- 



R F ^ 



R, 



Dividing out the R f term, we now have 



/r = 



R. 



1 + 



h PE (R -: R E ) 
Rr 



(5.29) 



This expression for the collector current can be compared to Eq. 5.12 
for the four-resistor bias circuit. We see that they are the same in form but 
differ in the stabilizing ratio, which now is 

/H Re 



S. R. - 



R, 



(5.30) 



The effect of this value of stabilizing ratio can be assessed by reference to 
Fig. 5.6. Since both R and R E are included in the numerator, the division of 
resistance between emitter and collector circuits has no effect on stability. 
Resistor R, can be determined from 

V CE = R,1 B + V BE 

'CK 'BE ~ VCE 

h m Ib 



R t - 



(5.31) 



It is usually possible to make the stabilizing ratio and the stability for the 
four-resistor bias network larger than for the voltage-feedback circuit since 
R f is relatively large. The voltage-feedback circuit with R, saves one re- 
sistor, however, and avoids the power dissipation due to the current /, in 
the R„ R 2 network. This point may be significant in battery-operated equip- 
ment. 



112 



DC Bias for the Transistor 



5.8 Design of a Voltage-Feedback 

Bias Circuit 

We shall use a transistor with the following values: / c = 2.3 raA, I B = 30 //A, 
V BB = 0.5 V, and h FB = 75 in a C-E circuit with V cc = 15 V. 
Making V CE = V cc \2 = 7.5 V, we find 



* + «—* 



15 



2 x 0.0023 



2 3200 Q 



With R E l c = 0.I5K CC = 2.25 V, we can calculate that 



ff _2.25 2.25 
£ : l c " 0.0023 

and 7? = 3200 - 1000 - 2200 Q. 
Using Eq. 5.31, 

d 'ce ~ 'BE 

Rf ~ T m 



siooon 



7.5 - 0. 5 
30 x 10 



i = 230,000 fi 



This completes the circuit design. 
The stabilizing ratio is 

S.R. = 



R + Rm 

3200 



= 0.014 



~ 230,000 

From Fig. 5.6 we see that the l c value will change about 275 per cent for a 
500 per cent change in lt FE . 

This figure for stability ratio is not so good as can be obtained for the 
four-resistor network, where we might have R„ s= 10,000 Q. Then 

bR - r- b Tom~ 

In the voltage-feedback circuit the value of R f is determined by base current 
and R f is large. The large R f value reduces S.R. as shown. 



5.9 Bias for the Emitter Follower 

The bias circuit for the emitter follower, Fig. 5.11, will now be considered. 
We start with the four-resistor network but make R — 0. Resistor R E now 
becomes the total load resistance. 



Bias for the Emitter Follower 



113 




(a) (h) 

Figure S.I I (a) Bias for the emilter follower; (b) design example. 



Using K cc /2 as the voltage across the load R e , we have 

Vcc 



K- = 



W, + lc) 



Vcc 
2I C 



and 



V E S RrIc = 



~F 



The voltage V B at the base of the transistor must be 

V B - V„ + V R 



(5.32) 



(5.33) 



(5.34) 



The input resistance of the emitter follower is very large and l„ will be very 
small. Then /, £ I 2 and the voltage-divider ratio for R, and R 2 will accu- 
rately determine V B . Then 



V "- R x rR l Vcc 



(5.35) 



We have only this single relationship with two unknown resistors. Again we 
must rely on experience. A value of 60,000 Q may be arbitrarily chosen for 
R : + R 2 , after which we can solve for R, and /? 2 separately. 

We can use Eq. 5. 12 for determining l c since that equation was ob- 
tained for the four-resistor bias network that we are designing but with 
R = 0. That is, 

hrs 



I -^ 



i+h FE q? 



R 



(5.36) 



114 



DC Bias for the Transistor 



Review Questions 



116 



and the stabilizing ratio is 



SR =fc 



Due to the large value employed for R E in the emitter follower, the 
stability ratio usually exceeds O.l. This ensures excellent stability of / c in 
the emitter follower circuit. 



5. 10 Design of the Emitter Follower 

Circuit 

Let us use a transistor with h FE — 100, V BE = 0.5 V, V cc = 20 V, and 
I c = 2.3 mA in an emitter follower (C-C) amplifier. 
We find R E from the g-point data. 



* £ - 2T 



20 



S 4300 CI 



, c 2 x 0.0023 

It follows that V E = V cc /2 = 20/2 = 10 V and by Eq. 5.34 we find V B as 

V B = 0.5 -t 10 = 10.5 V 
We choose R, I R t = 60,000 Q and can find R 2 from Eq. 5.35 as 



R 2 = (R>+ RJ 



Vb 



(5.37) 



Then 



= 60,000-!^ = 31 ,500 ^ 30,000 Q. 



R, - 60,000 - 30,000 = 30,000 Q 



The circuit design is complete. 

With /?, = R t = 30,000 Q, the value of R B is 

- /?,/?, _ 3 X 10* X 3 X 10* 

» ~ R,+R 2 ~ 6 X I0« 

= 15,000 Q 
The stabilizing ratio is 



SR . _**= 430 ° 



R„ 15,000 



= 0.287 



Figure 5.6 shows that, with this large ratio, excellent stability of I c is provided 
against changes in h FB . 

The complete circuit appears in Fig. 5.1 1(b). 



5.11 Comments 

We can now design complete amplifier circuits, including the necessary bias 
networks. These use only the source V cc , avoiding the cost of separate base 
and collector sources. Hopefully, these bias networks do not affect the gain 
or other performance figures of our amplifiers. The use of the emitter resis- 
tance R E often reduces the gain, however, and this must be allowed for in 
overall amplifier design. The use of the current-wasting resistors R, and R z 
introduces a cost in additional power. The effective value for /?, and R 2 , 
known as R B , shunts the input circuit of the transistor and wastes some of 
the input current, again reducing the gain. But the offsetting improvement in 
stability of the Q point and the elimination of individual amplifier adjust- 
ment in production are very significant. 

We have discovered that payment in some form is extracted elsewhere 
for every advantage gained in a circuit. 

Three basic bias circuits have been studied and the stability ratios are 
compared as 



S.R. 



Fixed bias 
Four-resistor bias 
Voltage feedback 



Iife (Eq. 5.I6) 

R B IR B (Eq. 5. 1 2) 

(* + **)/*/ (Eq. 5.30) 



Because of the flexibility in choice of R E and R B , the four-resistor bias net- 
work usually is the most stable form. 

REVIEW QUESTIONS 

5.1 What is meant by the Q point? What factors do we consider in locating the Q 
point? 

5.2 Give two reasons for h FE not being at the rated value. 

5.3 Why docs the collector power dissipation bound the operating region of a 
transistor? 

5.4 Why docs the maximum voltage bound the operating region of a transistor? 

5.5 What is meant by the saturation line of a transistor? 

5.6 What are the end points or intercepts of the dc load line? 

5.7 How is the slope of the load line related to the resistance in the output circuit ? 

5.8 Why must the Q point be located on the dc load line? 

5.9 Explain how an ac signal or I b value shifts the operating point along the load 
line. 






116 



DC Bias for the Transistor 



5.10 How docs / cflo vary with (cmpcrature? 

5.11 Name one advantage and one disadvantage of the fixed-bias circuit. 

5.12 What is meant by fixed bias? 

5.13 What is the stabilizing ratio of the four-resistor bias circuit? 

5.14 What range of I c variation could occur in a four-resistor bias circuit with S.R. 
= 0.05? 

5.15 How does the fixed-bias circuit stabilize against temperature-caused changes 
in V„1 

5.16 Name an advantage of the voltage-feedback bias circuit; name a disadvan- 
tage. 

5.17 How is the emitter voltage usually related to V cc in an emitter follower? 

5.18 What is the usual relation between V E and V cc in a C-E amplifier? 

5.19 From the load line, explain why we choose V CE as one-half of Vcc to start our 
circuit design. 

5.20 What is the stabilizing ratio for the voltage-feedback circuit? 

5.21 Name three forms of payment we make to obtain stability of the Q point. 

PROBLEMS 

5.1 For the transistor of Fig. 5.7, choose V cc = 20 V, / c - 4 mA, and V CB -- 9 V; 

draw the load line. What is the value of the load resistance? What is the base 

current 1„ at the Q point? 

# 

5.2 For the transistor with characteristics in Fig. 5.1, choose a Q point for 1 B = 
80 M and V CE = 12 V. 

(a) What / c is obtained? 

(b) With V cc = 25 V, what is the voltage across the load ? 

(c) What is the load resistance? 

5.3 The Q point at C in Fig. 5.1 is used. With V cc ^ 40 V, what load resistance is 
needed ? 

5.4 With the fixed-bias circuit in Fig. 5.4(a), find R and R B to place the Q point at 
C, Fig. 5.1, with V cc = 40 V. 

5.5 A fixed-bias circuit is designed as in Fig. 5.12(a). 

(a) With V BE = 0.5 V, V CE = Vcc/2, determine the (?-Point values of l B and 
/c 

(b) What is h PE of the transistor at the Q point? 

5.6 For the circuit in Fig. 5.12(b), 

(a) Determine R B to bring I c to 1.8 mA, ir/», E = 50. 

(b) What is the stability ratio for the circuit ? 

5.7 The transistor of Fig. 5.1 is used in the circuit in Fig. 5.13(a), with V cc -= 25 
V, V BE = 0.5 V, I c = 4.5 mA. Determine R, and R 2 values needed. What is 
the stabilizing ratio? What percentage will I c change if h VE changes from 40 
to 100? 



1 



Problems 



117 




>+ 10 V 



<( ° 




o II- 




-|( ° 



i kn 



(a) 



(b) 



+ 30 



« ° 




Figure 5.12 



«+25 




(a) 



(b) 



Figure 5.13 



5.8 With a Q point at l c = 3.5 mA, h, E = 60, Vcc = 25 V, and V BK = 0.5 V, 
determine the values of the other resistors needed for the circuit in Fig. 
5.13(b). What is the stabilizing ratio for the circuit? 

5.9 Determine the g-point currents and voltages for the circuit in Fig. 5.14(a), 
with V cc -\1 V, V BE = 0.6 V, and h fE = 55. 

5.10 Determine the bias currents and transistor voltages for the circuit in Fig. 
5.14(b), with V BE = 1-0.3 V and h PE = 75. Note: The circuit is an emitter 
follower. 

5.11 Determine the design values of the resistances for the circuit in Fig. 5.15(a), 
with V cc = 20 V, I c - 6.7 mA, and h FE = 60, using the usual circuit assump- 
tions. Determine V CE and V E and the stabilizing ratio. 



118 



DC Bias for the Transistor 



Problems 



119 





° It 



250 k 



(a) 



+ 20 V 



fb) 



Figure 5.14 




(a) 




Figure 5.15 



5.12 Lei R E = in Fig. 5.13(b) and calculate the bias currents and transistor volt- 
ages for R, - 150 kSl, R - 2.1 kSl, y cc = 9\, V BE = 0.6 V, and h FE = 40. 

5.13 Using the fixed-bias circuit in Fig. 5.4(a), 

(a) Determine the collector-emitter bias voltage (V CE ) for R B = 250 kSl, R = 
2 kQ, K cc = 12 V, V BE = 0.3 V, and h fE = 64. 

(b) Find l c and / fl . 

5.14 For the circuit in Fig. 5.13(b), with R, - 250 k£2, rt - 2 k£2, R E = 500 kfi, 
Kx: = 9 V, V BE --= 0.5 V, and h FE = 55, 

(a) Find the dc currents and V CE . 

(b) What is the stabilization ratio for the circuit? 






5.15 For the circuit in Fig. 5.13(b). find V CE for R, 47 kfi, R E =- 750 Q, R = 
500 SI, V BE 0.5 V. h n = 55, and V cc = 18 V. 

5.16 What value of emitter resistor would be needed to make the stabilization ratio 
equal 0.02 for the circuit in Fig. 5.13(b), with R = 750 SI. /?, = 50,000 ft. 
and h rS - 40? 

5.17 Calculate the bias currents and transistor voltages for a four-resistor bias C-E 
amplifier with R, - 56,000 SI, R 2 = 5000 SI, R E = 750 Si, R - 6800 0. 
V cc - 24 V, V BE 0.6 V, and h, E = 50. 

5.18 Design an amplifier circuit as in Fig. 5.13(b) for a transistor having h tE = 45, 
I c = 5 mA, V C c = 20 V, and V BE - 0.5 V. Make the usual circuit assump- 
tions. 

5.19 For the emitter follower circuit in Fig. 5.15(b), find the value of Vg, for R B = 
90 kfi, R E 1500 S\, V cc = 25 V, V BE = 0.35 V, and h rE -- 60. Also find 
the stabilizing factor and predict the I c change tf h Fl . changes to 100. 

5.20 Design an emitter follower, using the circuit in Fig. 5.15(a) with R = 0. The 
transistor has li FE = 100, V BE - 0.5 V and V cc - 15 V, l c - 2.5 mA. Make 
the usual assumptions. 




6 



The Field-Effect 
Transistor 



The field-effect transistor (FET) was an early proposal by Shockley that 
had to wait for the development of new production methods to become a 
practical device. It is a unipolar device because the current is either of holes 
or of electrons, in contrast to the bipolar transistor of pnp or npn form. The 
cross section and consequently the resistance of the conducting path in this 
device may be controlled by a signal voltage applied to a gate electrode. 

With a high input resistance of 100 Mfi or more and generating less noise 
than a bipolar type, the FET is well suited as an input amplifier with low- 
level signals. 



6. 1 The Junction Field-Effect 

Transistor 

Figure 6.1(a) illustrates the operating principle of a junction field-effe • tran- 
sistor (JFET). The thin slab of semiconductor has contacts at each <.. d, a 
source S for the mobile charges, and a drain D for extraction of the charges. 
A p electrode forms a junction on the /; wafer and is known as the gate G; 
the thin region under the gate is called the channel. If n material is chosen for 
the channel, the conduction is by electrons. Ifp material is used, the conduc- 
tion will be by holes. 

For discussion, we shall use n material for the channel. With a voltage 

720 






The Junction Field-Effect Transistor 



Depletion 
Region 



121 



Depletion 
Region 




Pincheil-OI'l 
Channel 




& 



H|lt 



(a) (b) 

Figure 6.1 (a) Low voltage applied to the FET; (b) in the pinch-ofT condition. 

V DS applied, electrons move in the channel from the source to the drain. 
There is a progressive voltage drop along the bar and, with the source and gate 
being connected, points in the channel will be positive to the source and gate. 
With a positive n channel and a negative p gate, the junction at the gate- 
channel interface is reverse-biased. As a result, a depletion region forms in 
the channel as show/i. This depletion region appears largely in the /; channel 
because the /?-gate material is more heavily doped. 

The voltage drop is distributed along the gate and the reverse voltage at 
the right end of the gate is larger than at the left end. Because of the greater 
reverse voltage, the depletion region at the right end of the gate is thickened, 
as shown. The depletion region has no mobile charges and the channel current 
is confined to the wedge-shaped portion of the channel. 

The drain current increases as V DS is raised, resulting in the curve of (1) 
in Fig. 6.2(a). As there is greater reverse voltage between gate and channel, 
the depletion region begins to pinch off the channel as in Fig. 6.1(b) and the 
rate of current increase falls off to (2) on the curve. With further increase 
in V os the pinched-off region lengthens and the current curve flattens to (3). 
This region of essentially constant current, relatively independent of V DS , is 
called the pinch-off region. 

Placing the gate at — I V to the source 5 further reverse-biases the junc- 
tion and drives the depletion region into the channel. Saturation of the 
current occurs at lower values of V DS and at lower currents. In the pinch-off 
region the current is sensitive to the gate-source voltage V os ; as such, we 
have characteristics suited to the use of the FET as a control device, shown 
in Fig. 6.2(b). 

The operation is said to be in the depletion mode, with the increased 



122 



The Field-Effect Trensistor 





(ii) 



(b) 



Figure 6.2 (a) Current variation with V D s, with gates connected to source; (b)an 
output curve family, showing l D controlled by V s- 

negative gate voltage depleting the channel of charges and lowering the 
channel current. 

The abrupt rise in current above (3) in Fig. 6.2(a) is due to an avalanche- 
type voltage breakdown in the depletion region between gale and drain and 
this places an upper limit on the gale-drain voltage. The breakdown voltage 
is designated BV DOO , the source being open-circuited. 

The input resistance is that of the gate-channel reverse-biased junction. 
The gate area is small so that the leakage resistance is of the order of 100 MQ. 
A depletion capacitance C us appears in parallel. This capacitance is that of 
the reverse-biased junction, due to the dielectric effect of the depletion region 
between the gate electrode and the channel. This capacitance may be in the 
range of 2 to 10 pF. 

Currents of milliampere size in the channel can be controlled by the gate 
voltage. With the input current from the gate extremely small, we obtain 
significant amplification in a circuit such as that in Fig. 6.3. 




Figure 6.3 The JFET in the common-source amplifier. 







The MOS Field-effect Transistor 123 



6.2 The MOS Field-Effect Transistor 

The metal-oxide-semiconductor FET (MOSFET) has its gate electrode insu- 
lated from the channel by a very thin layer of silicon dioxide. A wafer of high- 
resistivity p silicon is used and n impurity is diffused into a region near the 
top surface to form an n channel of moderate resistivity. Low-resistance n 
contacts arc diffused through a mask at the channel ends as source and drain. 
The surface is covered with a thin layer of silicon dioxide and a small metal 
gate electrode deposited over the channel. The thickness of the insulating 
layer is usually less than 10" 3 mm. 

Application of a negative bias to the gate drives electrons from the n 
channel immediately under the gate, depleting the region of free charges. 
The thickness of the depletion region varies with gate bias and constricts the 
conduction area in the channel. In Fig. 6.4(a) the condition at pinch-off is 
shown. 

The drain current is large with zero gate voltage and is reduced as the gate 
is made more negative. This is the depletion mode of operation; a family of 
output characteristics is shown in Fig. 6.5(a). 



Metal 




Depleted n Channel 

p Region 

(b) 




Figure 6.4 (a) Depletion-mode MOSFET at pinch-off; (b)channel configuration 
of enhancement mode; (c) transfer curve for an n-channel enhancement-mode 
transistor. 



124 



< 
E 



< 

a 



The Field -Effect Transistor 




10 




^— ^~~C= + 6 


8 




— -** +5 


6 




+4 


■1 




+3 


2 




— + 2 




- 1 — "; 



12 16 20 



(b) 

F/ffure d.5 (a) MOSFET output characteristics: depletion mode, n channel; 
(b) enhancement mode, p channel. 

When constructed with a p channel, a positive voltage on the gate repels 
holes from the wafer and a depletion region forms under the gate, as indi- 
cated in Fig. 6.4(b). There is no current in the channel. If the gate is made 
more positive than some threshold V T , however, the positive gate charge 
attracts electrons from the negative source and builds an // channel between 
source and drain. This is shown in black in the figure. Current now passes 









The Load Line for the FET 



125 



in this channel, with a transfer curve between voltage and current as shown 
in Fig. 6.4(c). 

The positive drain potential readily sweeps out electrons, the channel is 
narrowed at the drain end, and pinch-off occurs. With zero current at zero 
V s and an increasing drain current with increasing gate voltage, we have 
the enhancement mode of FET operation. 

The input resistance of the MOSFET is due to the silicon dioxide layer 
and can be as high as 10 9 to 10" Q. The gate has a length of about 15 X 
10" 3 mm and the capacitance between gate and channel is about 1 to 4 pF. 

MOSFET construction is particularly adaptable to the processes of impu- 
rity diffusion and metal deposition that arc carried out in the production 
of integrated circuits. 

6.3 Symbols for the FET 

In Fig. 6.6 we show the several circuit symbols used to identify field-effect 
transistors. In Fig. 6.6(a) and (b) the arrows show the materials used in a 
manner similar to the diode symbol; at (a) we have a p gate to an //channel 
and at (b) there is an n gate on a p channel. The arrows tell us the polarities 
of the needed bias voltages. The gate electrode identifies the source by being 
placed above 5. 



J$-o 






S v - — " D S v — ^ D S v -4--' D 

Wafer Wafer 

(a) (b) <c) (d) 

Figure 6.6 JFET: (a) n channel; (b) p channel. MOSFET: (c) //-channel deple- 
tion; (d) p-channel enhancement. 

The isolated gate symbol in Fig. 6.6(c) and (d) identifies the transistor as a 
MOSFET. The arrows in the wafer leads indicate the materials, again follow- 
ing the diode symbolism; at (c) we have an //-channel depletion mode and 
at (d) there is a //-channel enhancement-mode transistor. 

6.4 The Load Line for the FET 

We may draw a dc load line on the characteristics of an //-channel JFET in 
Fig. 6.7, as done for the bipolar transistor in Sec. 4.7. The circuit in Fig. 
6.7(b) is a common-source circuit. We write from the output circuit 

v DS - V DD - Ri D (6.1) 



126 



The Field-Effect Transistor 



1 



























— 1 


v« = 









(?_?N 












1 














Q 
? 








_ 2 








-v 








3 

-4 

S 








£s 







10 

Yds 

(a) 



v ds v dd 



15 



20 



lo 1(- 



|o- 



H( 02 




'•R, 



r cc 



'l)S 



(b) 



-02 



Figure 6.7 (a) Depletion-mode, JFET characteristics; (b)common-sourcecircuit. 

which represents a straight line. This dc load line can be drawn for an .v-axis 
intercept at V DD and a j'-axis intercept at i D = V DD /R. These quantities 
should be so chosen that the load line traverses a region of uniform transistor 
characteristics. 

The Q point or zero V, point must lie on the load line for the circuit. A 
signal applied to the gate will vary v as and cause the operating point to move 
from Q up and down along the load line. If the output waveform is to be an 
undistorted image of the input wave, then the Q point must lie in a region of 
uniformly spaced output curves. The curve for v as — serves as one limit for 
the operating region. Point A at that limit should be in the saturation portion 
of the current curve. Voltage V DD should be less than the maximum rated 
drain-source voltage for the transistor. 



Obtaining Bias lor the FET 



127 



With these limitations, point Q of the figure seems a reasonable choice 
and a dc load line can be drawn through Q and V OD as an intercept. 

Wc can determine the value of load used as the reciprocal of the slope of 
the load line. The slope can be measured on the triangle formed by the load 
line and the axes as 

A/ V DD IR __ 1 
V nn R 



slope = 



AVr 



(6.2) 



Wc then have 



R--= 



12-0 
0.0034 - 



2= 3500 « 



We use the output characteristics to choose a Q point for bias determina- 
tion for the transistor. The (?-point location gives us the quantities I D , V as , 
and V DS . 



6.5 Obtaining Bias for the FET 

The channel-resistance and the gate-control properties of FETs are subject 
to considerable variation among units of the same type. In addition there 
are changes due to temperature. Bias circuits must be designed so that some 
self-regulation of the Q-point drain current is obtained. The circuit should be 
able to compensate when one transistor is substituted for another of the 
same type but with differing parameters. 

The problem is illustrated in Fig. 6.8 where the drain-current versus gate- 
voltage transfer curves are plotted for three transistors of the same type. The 
zero bias-drain current I DSS varies from 25 mA for the high unit to 6 mA 
for the low unit. With fixed-voltage bias, V^ ■= —1.2 V, the drain current of 
the high unit would be 12 mA and that of the low unit would be zero, or the 
transistor would be in a cutoff condition. Fixed-gate bias is not a satisfactory 
solution to the problem. 

Self-bias, as shown in Fig. 6.9(a), is preferred over fixed bias. This elimi- 
nates one voltage source at the cost of two resistors. Resistor R, is large, 
usually 1 Mfl or more, and is present only to fix the average gate voltage at 
ground. This is accomplished since /„ = and there is no voltage drop in 
R,. The indicated polarities in Fig. 6.9(a) lead to an input circuit equation of 

y os + y* = o 

-V 0S =V S = R S 1„ (6.3) 

The negative sign shows that the gate is negative in relation to the source. If 
I„ increases, then V s rises and the gate is made more negative. But a more 
negative gate tends to reduce I D toward its initial value. Self-regulation of 
I D is thus accomplished by this circuit. The amount of stabilization of /„ 



128 



The Field-Effect Transistor 






15 



20 



< 
S 15 



10 









l DSS 


- 




fl 




- 


VCG 






- 


1 
\ I 

/! 


si 




- 


/ 1 

/ i 

/ i 

/ i 


^ i 








o / 




/l 


/ \ 

/ i 

/\ ! 


-h 


1 



-3 



-2 



- 1 



+ 1 



Figure 6.8 Drain current versus gate voltage for several 3NI28 FETs. 

is proportional to #„ however; large values of R s require increased V DD and 
also result in a reduction of gain. 

We may decide that about 15 per cent of V DD should appear across R s 
at 2-P°i nl current. Then the resistor voltage will be 

V S ~0A5V DD (6.4) 

Due to the flatness of the V os curve, we can read /„ and can find R s as 



Rs = ¥ (A) 



(6.5) 



The voltage drop across the transistor may be made equal to V DD \2 at the Q 
point and so we know that 



ft + J^S^p (fl) 



(6.6) 



Then we can determine the value of the load R. The load line for R + R s 
can be drawn from the x intercept at V DD to the y intercept at V DD I(R + R s ). 
The Q point and V DS are fixed by the intersection of the V os line and the dc 
load line. 

This completes the design of the self-bias circuit. 

The input and output capacitors are blocking capacitors, as before. Since 
they are expected to represent negligibly small reactances, they will not be 
considered at this time. 



The Four-Resistor Bias Network 



129 



10 




< 
E 



KnO—H 



/ 




••«• = o 


v — 






L 






- 1 








< 






- 2 


r 


S 






\ 




Q 


3 


r 








I 




X N 


S," 


r 






N 



(a) 



10 



r DS 



(b) 



IS 



20 



Figure 6.9 (a) Self-bias FET amplifier; R + R, = 4300 fi; (b) characteristics 
and load line. 



6.6 The Four-Resistor Bias Network 

The circuit in Fig. 6.10(a) requires one more resistor than the self-bias circuit 
but gives greater design flexibility in that the gate bias can be negative, zero, 
or positive as required. We choose a Q point in Fig. 6. 10(b) and find /„. Using 
the arbitrary rule of Eq. 6.6, 



R-\ R s = 



In 



(fl) 



(6.7) 



The voltage across R s may be arbitrarily chosen as 20 per cent of the total 
drop across R and R s so that 



r s i d ~ °^j»> = o.io y B 



(6.8) 



With R s determined, we can find the load R. 

In the circuit we show V os with gate positive to the source. If the gate is 
to be negative, then K c . v will carry a negative sign. Around the input loop 



V -V os -V s = 

Vo = Vos + V a 



(6.9) 



130 



The Field-Eflect Transistor 



"b.o_|(- 




20 



IS 



= 



e to 






{ 




Vgs=0 






-1 


h 


Q 




- 2 


!/ 


\ 






y 


\ 




3 


X 


-A 


-5 


-4 

i 



10 20 30 40 

(a) (b) 

Figure 6.10 (a) Four-resistor bias circuit ; (b) n-channel depletion FET, 3N 1 39. 

Since /„ = 0, we can use the voltage-divider relation to determine /?, and R 2 . 
That is, 



^ = 



/?, + /? 2 



' DD 



(6.10) 



Again we have two unknowns with only one equation. We arbitrarily select 
R, + R 2 to draw a small current from V DD , compared to /„. Having R t ) R 2 
assumed, we can solve Eq. 6.10 for R 2 and then obtain /?,. 

This completes the amplifier design. It may seem that much of this pro- 
cedure is arbitrary but we remember that transistor curve families are drawn 
for average units and differences between transistors will create circuit varia- 
tions that make more precise circuit design unnecessary. The self-regulating 
action of the circuit will usually bring the Q point to within about 10 per cent 
of the expected I D value for normal transistor and resistor variations. 



6.7 Design Examples 

Example 1: Using the FET represented by the curves in Fig. 6.9(b) with 
a e-point current I D = 3 mA, with V DD - 20 V, find the exact Q-point 
location and the design values for R, R„ and R s for the self-bias circuit of 
Fig. 6.9(a). 

Selection of I D = 3 mA places our Q point somewhere along the V os ■= 



Design Examples 



131 



-3 V characteristic. But — V os = V s = 3 V and, using Eq. 6.5, we find 

^ = ^ = D^3= 100 ° Q 
Then using (/? ■ R S )I D = K DD /2, we have from Eq. 6.6 



R + R s = ^? 



20 



= 3300 Q 



2(0.003) 

The load line of the figure is drawn between V DD — 20 V and a y intercept 
at i D = Y DD KR ->- R s ) = 6 mA. The Q point is at /„ = 3 mA, V os = -3 V, 
and V DS = 10 V. 

Arbitrarily we select R, = 1 MSI so the circuit resistances are 

r s = looo n 

R = 2300 fi 
R, - 1 Mfi 

Example 2: The circuit in Fig. 6.10(a) is supplied with a transistor having 
the characteristics in Fig. 6.10(b). Supply voltage is V DD = 20 V and the 
V os = —2 V line is considered a suitable location for the Q point of this 
/i-channci depletion FET. 

We shall choose (R -| R S )I D = ^dd/2 and this assumption fixes V DS = 
V DD /2 so that our Q point is located at V as = -2 V, V DS = ^ = 10 V. 
A dc load line for R ■ R s \s drawn from the .v intercept at V DD through the 
Q point. 

The Q point fixes I D = 6.5 mA. Then we have 



* + *, = !&.« 20 



= i5oon 



2/ D "0.013 

We choose the voltage V s as 15 per cent of V DD or 

^ = ^/ D = 0.15K oo 
0.0065/^.-0.15 x 20 = 3V 

*^O565 = 470fi 

after rounding the resistance to the nearest standard value. Then we have 

R- 1500- 470 Q: 1000 Q 

We can design the voltage divider to give us the needed value of V c from 

V = V s + V as 

= 3.0 + (-2)= 1.0 V 

That is, the source is 3.0 V positive to ground and the gate is 1.0 V positive 




132 



The Field-Effect Transistor 



to ground. Thus the gate is -2 V to the source as we determined from the 
Q point. 

Since current through R, + R z results in a power loss, we choose R, + 
R 2 to take a current that is small with respect to I D . Selecting /, k 300 /zA, 
we have 



Rt+Rz = 



20 _ 



0.0003 ~ 0.0003 



~~ 67,000 n 



Because / = 0, the voltage-divider ratio will be accurate in setting V c and 
we have 

^2 1/ 



Vn = 



R, +/? 2 



V 



R^(R l+ R 2 )^3_ 

'DO 

= 67,000^ = 3300 Q 

Then R, = 63,700 n. 

As the nearest standard resistance values, our circuit will employ 

r=\oooq /?, = 330on 

R s = 470 Q R z - 62,000 Q 



6.8 The FET as an Amplifier 

As with the pnp or n/w bipolar transistors, the FET has three internal elec- 
trodes. We want to connect the FET, however, to four circuit terminals or 
two ports. Some FETs have four leads but one is a second gate that is 
largely used for bias control. Here our gate symbol indicates the ac signal 
or control element of the FET. 

The three choices of common lead provide three amplifying circuits, the 
common-source circuit, the common-gate circuit, and the common-drain 
circuit. The common-drain circuit is also known as a source follower and 
serves the same function as the emitter follower. In the common-gate circuit 
the high input resistance of the FET is lost. 

Depletion-mode operation is more commonly used for amplifiers, while 
the enhancement mode is used mainly for switching purposes in digital cir- 
cuits. 

6.9 Circuit Characteristics of the FET 

The output curves for the FET predict the operation of the devices when used 
in an amplifier circuit. The JFET and MOSFET curves have the same general 
shape and as a result the devices may be discussed together. We choose the 






Circuit Characteristics of the FET 



133 



source as common and the voltages are measured to that electrode, as 
reference. 

As an amplifier, the FET is operated in the pinch-off condition where 
variations in drain current arc dependent almost entirely on gate voltage. 
The slope of the output curves is 

^£- B -L (6.11) 

At; DS r d 

where r d is called the drain resistance of the FET. The low and nearly con- 
stant slope of the curves in the pinch-off region shows that r d is high and 
constant, almost independent of V os . This is a characteristic of a constant- 
current generator. 

The input resistance r os is considered very large and usually neglected. 
The effect of the shunting capacitance will be considered in a later chapter. 

A typical FET transfer curve, relating the drain current /„ to the input 
signal voltage v as , is drawn in Fig. 6. 11(a). This curve is found to be predicted 
by 

« D =/ D «(l-^) 2 (6-12) 

where f DSS is the value of /'„ with the gate shorted to the source S; this current 
is indicated by the small circle on the ordinate in Fig. 6.11(a). The constant 
V p is the pinch-off value of v DS , as at (2) of Fig. 6.2. It is found from a pro- 
jection of the slope of the transfer curve from / DSS , which intersects the 
abscissa at a voltage of V„/2. 

A useful transistor performance figure is the transconductance g m . This 
is the change in output current per volt change in input voltage. This is 
found from the slope of the transfer curve as 



^- = ^r < mhos) 



(6.13) 



The slope is a variable and g m varies with i D . The extent of this change for a 
typical FET is plotted in Fig. 6.1 1(b). The selection of the 0-point current 
then determines the value of g m being used for a given FET and circuit. 

Equal changes in v as do not produce equal changes in /„ on the transfer 
curve because of the curve variation. If we reason that a curve is made up of 
short straight lines and if we choose a small segment of the curve, however, 
we can assume that segment is straight. Translated to the FET, this means 
that wc can use the FET for relatively distortionless amplification if we apply 
only small-signal voltages, causing i D to vary over only a small segment of 
the transfer curve; that is, we might restrict the input signal to ±0.2 V peak- 
to-pcak. Thus the FET is most often employed as a linear amplifier in the 
input, or small-signal, stages of a system. It is also useful in digital-switching 
circuits, where amplitude distortion is not important. 



1 



The Equivalent Circuit of the FET 

































































































/i 


















/ / 


f 


















/ 1 

f 1 

1 




















r 
i 
































1— d- 


1 











'OSS 



i„ (mA) 



- S 



3 V ± -2 

2 

(a) 



of 



10,000 
8000 
6000 
4000 
2000 






































/ 












/ 












/ 













2 4 6 8 10 12 

in (mA) 

<h) 

Figure 6.11 (a) FET transfer curve, depletion-mode FET; (b) variation of gm 
with i'd for a 3N128. 



134 



Typically, wc would expect r t to be in the range of 5000 to 50,000 SI for 
MOSFETs and above 100,000 fi for JFETs. The value of g m may be expected 
in the range of 1000 to 10,000 //mhos. This latter figure means 

0.0I0 = A 
"«« 

and indicates that we shall have an output current of 10 mA ac for each ac 

volt input. 

6. 70 The Equivalent Circuit of the FET 

Figure 6.12(a) shows the internal elements of an FET arranged as a two-port 
network in common-source connection. Too complicated to be readily 
analyzed, we reduce it to the simple equivalent circuit at Fig. 6. 12(b) by con- 
sideration of the relative magnitudes of the resistances and reactances. The 
equivalent circuit that is developed is satisfactory for applications not exceed- 
ing frequencies of a few megahertz. 

The element r„ is the reverse-biased junction resistance of the J FET or 
the silicon dioxide layer resistance of the MOSFET. These resistances are so 
high that we may consider r f , as representing an open circuit. 

The series circuit of C c and r c represents the capacitance between gale and 
channel and the series resistance of the channel. Being only a few picofarads 
and a few hundred ohms, their effect is negligible at low radio frequencies 
and they are eliminated from the equivalent circuit. 

Capacitance C, d is the gale-to-drain capacitance and includes the capaci- 
tance of the transistor mounting; being only 1 to 3 pF and of very high 
reactance we drop the element from present consideration. 

It was pointed out that the output current of an FET behaves as that of a 
current generator and so the internal generator is shown as the current source 





-o2 I °—\ — • 



Q*m V K> 



,/d 



V » 



©a-"* 



I 



-©2 |c~ 






-o2 



r a\ I 



III 



-i AS _ 



-U2 



/ 



(a) (b) 

Figure 6.12 (a) The internal FET circuit; (b) equivalent circuit for the FET. 



1 



136 



The Field-Effect Transistor 



Thus in Fig. 6.12(b) we have the same form of equivalent circuit as we 
previously used for the bipolar transistor; however, the input circuit of the 
FET takes zero current. The circuit elements are assumed constant around 
a given Q point and this restricts our operation with ac signals to a 
linear region of the output characteristics and to small input signals. Since 
the bias sources are not included, the circuit is useful for analysis with ac 
signals only. 

The effects of C c and r c and C, d at higher frequencies will be discussed later. 



6.11 The Common-Source Amplifier 

In the common-source amplifier in Fig. 6.13(a), the input signal V, — V 
is applied between gate and source and the output V a is obtained between 
drain and source. In Fig. 6.13(b) we have replaced the FET between G,S, 




o2 



lo— 1( 1 


\Kc. 


S 


r e 


Vi 








1 


■- Vac 












o 




(a) 



(b) 



Figure 6.13 (a) Common-source FET amplifier; (b) equivalent circuit for (a). 

and D with its equivalent circuit consisting of a current source and the inter- 
nal drain resistance r d . Resistor R is usually so large that it appears as an 
open circuit. Except for the zero input current, the FET common-source 
amplifier is similar to the C-E circuit for the bipolar transistor. 

Voltage Gain 



The parallel effect of r d and the load R is represented by 

R - r " R 



(6.14) 



The Common-Drain Circuit 



137 



The current g„V„ produces a voltage V across the output as 



'o — 8m'^gt"f 8m 



r,R 



y,. 



• r d - r - R 

The negative sign results from defining the output voltage and current such 
that V„ = —Rl d - The negative sign indicates that a phase reversal has 
occurred between V„ and V B . 

The voltage gain in the circuit is obtained by dividing by V,„ leading to 



' V„ g "r d \ R 



(6.15) 



Another useful form of this expression can be obtained by dividing numera- 
tor and denominator by r d , giving 



A.= 



' + f 



(6.16) 



Usually R<g.r d and R/r d becomes small with respect to one and we then have 

A v ~-g m R (6.17) 

which is the same result as obtained for the C-E transistor amplifier. 

The voltage gain depends on the selection of a transistor with a high 
g„ and on the resistance chosen for R. 

Output Resistance 

The output resistance at the 2,2 port is defined as the resistance at that port 
with the input signal set to zero. 

With V, — V t , = the current of the current source is zero, a condition 
reached by opening the circuit of that source. The resistance remaining at 
the 2,2 terminals is r d and that becomes the output resistance of the amplifier. 

The value of r d is nominally high, perhaps 50,000 Q, and the gain 
approximates A v = —g m R- The circuit performance is similar to that of the 
C-E amplifier, except for the negligibly small input current taken by the FET. 



6. 12 The Common-Drain Circuit 

For the FET common-drain circuit in Fig. 6.14(a), the input signal is applied 
between gale and the drain and the output is taken between source and drain. 
As noted previously, the circuit is known as a source follower. This is because 
the output voltage V„ is approximately equal lo and varies with the input 
V,. In this, the circuit performance parallels that of the emitter follower. 

We replace the FET with its equivalent circuit in Fig. 6.14(b). For better 
understanding, we fold down the upper half of the output circuit about the 



138 



The Field-Effect Transistor 




lo 




(a) 



(b) 




Figure 6.14 (a) Common-drain circuit; (b) and (c) equivalent circuits. 

line of the source and in (c) have resistors r d and R s in parallel; inspection 
of (b) shows this to be correct. The current source appears to have turned 
upside down but its outgoing terminal remains connected to S. 

Voltage Gain 

Using Eq. 6.14 for the parallel resistance of r d and R s , the output voltage 



is 



y. = g m R.v„ - g m 



*Vf*J y 



(6.18) 



'r d + R s '" 

In this circuit the input signal is connected between gate and drain and from 
the circuit 

V, = V t , + V. 
V„=y,-K (6.19) 






The Common-Drain Circuit 



139 



(6.20) 



We make this substitution for V t , in Eq. 6. 18 and rearrange the terms so 

V (\ . 8™ r *R-i \ — o _JjRs_y 
°\ r d -, R s ) ~ gm r d ! R/> 

Multiplying by r d + R s and dividing by K„ the voltage gain from the input 
at 1,1 to the output voltage across R s is 

gm^Rs g„R s 

(6.21) 



A -&~ 

' V, r t + R a + gjjl, l+ Rs +gmRs 



The drain resistance is usually large with respect ioR s and R s /r d will be small 
with respect to the unity term. The voltage gain is then approximated by 

This result would be less than but near unity since g m R s is usually larger than 
one. There is no phase shift between the input and output voltages in this 
amplifier. 

Equation 6.22 has the same form as obtained for the emitter follower. 
Since the equivalent circuit for the amplifier is the same, we should expect 
this result. 

Output Resistance 

As a demonstration of the method, we shall find the output resistance 
of the transistor at 2,2, with the load R s open, by use of the impedance rela- 
tion for a voltage-source equivalent circuit 



R' = /?„ = If. 
Written without R s , Eq. 6.20 gives for V„ 



K„ = 



(6.23) 



(6.24) 



1 + g m r* 

With a short circuit across 2,2, we have V„ — and Eq. 6.19 gives V t , = V,. 
The current in the short circuit is 

l. c - g m V„ = g m V, (6.25) 

Dividing Eq. 6.24 by 6.25, we have R as 

g^y, ./ 

R , \+g m r/' _ r d (6.26) 

g m y, 1 + g m r d 

Usually g„r d » 1 and this equation reduces to a simple form, 



Sm 



(6.27) 



140 



The Field-Eltect Transistor 



Since g m is in the range of 0.00I to 0.01 mho, the output resistance of the 
circuit is small. 

Therefore the source follower, with only unity voltage gain, serves mainly 
as an impedance transformer, useful in amplification of a signal from a high- 
resistance input to a low-resistance output, such as a relay or a telephone line. 

The design of the bias circuit for a common-drain amplifier follows the 
procedure for the four-resistor circuit of Sec. 6.6 with R — 0. 

Example: Consider a source follower using a JFET with r d —- 50,000 Q 
and g„ = 0.0025 mho, as data given by the manufacturer of the JFET. Find 
the gain and output resistance with R s ^ 5000 £2. 

Using Eq. 6.2 1, we have 

g n Rs 



A.= 



1+7* + *.*, 



0.0025 X 5000 



1 + sgno + 00025 x 500 ° 



12.5 



12.5 



= 0.920 



1 +0.1 

If we had neglected the R s !r d term in the denominator, as in Eq. 6.22. we 
would have 

and the difference is negligible. 
The output resistance is 



r ~-L-_ !_ 

K °-g„~ 0.0025 



= 400Q 



6. 13 Design Example 

We shall perform the design calculation for a common-drain amplifier using 
the circuit in Fig. 6.10, with R = 0. The Q point is set at I D — 6.5 mA with 
y DU = 20 V, as in that figure. 

Using the arbitrary rule of Eq. 6.7, 



Rsln = 



and so 



D * DP 



20 
0.013 



^ 1500 Q 






Design Example 



The voltage V s is 



141 



V s = R S I D - 1500 x 0.0065 - 9.75 V 



This value is not quite the specified value of 10 V because of the rounding 
of the R s value. 
At the Q point, 

FW - -2 V 
and so 

Vo - Vos + Vs = (-2) + 9.75 = 7.75 V 

positive to ground. 

Since the gate current is zero, then resistors R, and R 2 serve only as a 
voltage divider for the gate voltage and we can arbitrarily choose /?, + R 2 
= I Mfi. Then Eq. 6.10 gives 

R 2 = (/?, + /? 2 )ik = 10 s X 2^ = 387,500 Q 



= 0.39MQ 



Then 



R, = I0« - 0.39 X 10* = 0.62 MQ 

again choosing a standard resistor value. 
The circuit is complete with 

R s ^ 1500 n 
R t =-- 0.62 MQ 
R 2 - 0.39 MQ 

The value of g m at the Q point can be determined graphically by measuring 
Ai' . v as 1 V, from the 1-V curve to the — 2-V curve, and reading the resul- 
tant A/' D along the vertical line for v DS — 10 V, through the Q point. The 
value of A/„ is read as 3.5 mA and we have 



*, 



_ A/„ _ 0-0035 _ 



0.0035 mho 



Aw os 1 

The value of r d is large, as can be seen from the flatness of the v os = — 2 V 
curve at the Q point and we can calculate the voltage gain from Eq. 6.22: 

g m R s 0.0035 X 1500 

t-*.** 1 + (0.0035 x 1500) 
5.25 



A v = 



1 + 5.25 



= 0.84 



The output resistance is 



R.^ = 



1 



*„~ 0.0035 



= 286Q 



142 



The Field-EI/ecl Transistor 



6. 14 Using the FET as a Variable 

Resistance 

When operated near point (I) of the curve in Fig. 6.2, the drain-source chan- 
nel of an FET can be used as a voltage-variable resistance (VVR). The device 
represents a resistance of several thousand ohms, as shown in Fig. 6. 15. 
Currents will be of microampere order and voltages will be a few hundred 
millivolts. 

The device has some application in signal level control and expansion. 
The basic elements of a volume control application appear in Fig. 6.15(a), 



o It- 



-v. 



GST 



<6 



G 



— o 






















IS 000 














/ 


















/ 




i 0.000 


































%nnn 






































— | — 















(a) 



4 
Vgs 

(b) 



-6 



-8 



Figure 6.15 (a) The FET as a voltage-variable resistance ; (b) resistance variation 
obtained. 

where the voltage across R is the output, as a varying fraction of V,. The 
gate voltage is varied by a rectified voltage derived from the input signal; 
as this increases, the FET resistance increases, passing on less of the input 
signal through the voltage-divider action of the circuit. The output signal is 
then controlled to a more constant level. 



6.15 Summary 

The unipolar FET, with either n- or ^-conduction properties, controls the 
output current magnitude by means of a voltage applied to a gate electrode. 
Its most important feature is the very high value of input resistance. It is well 
adapted to production by methods of diffusion and metal deposition. 
Because of the zero input current, the design of its bias circuits is simpler 












Problems 



143 



than for the bipolar transistor but the methods of circuit design are the same. 
We also find the FET useful in the common-source and source follower cir- 
cuits, which parallel the C-E and emitter follower circuits in performance 
and applicability. 

Because of its curved transfer curve, the FET can give linear amplifica- 
tion only with small signals. This makes it most adaptable to input circuits. 



REVIEW QUESTIONS 

6.1 Why is the FET called a unipolar transistor? 

6.2 Describe how pinch-off is obtained in an //-channel JFET. 

6.3 What is the construction difference between a JFET and a MOSFET? 

6.4 Explain how a depletion-mode MOSFET operates. 

6.5 What is meant by "enhancement mode"? 

6.6 What is meant by "depletion mode"? 

6.7 What is the significance of the arrow on the gate of a JFET symbol? Of the 
arrow on the MOSFET symbol ? 

6.8 What range of resistances can we expect in the gate-to-channel path of a 
MOSFET? 

6.9 What is the reason for limiting V uo for an FET? 

6.10 Why do we limit FET operation to small signals? 

6.11 The transistor in Fig. 6.7(a) has K oulm ,„ = 20 V. Bound the region for accept- 
able Q-point location for a linear amplifier. 

6.12 Relate r d and g m to the characteristic curves of an FET. 

6.13 Why is self-regulation of the drain current important in the common-source 
circuit? 

6.14 Compare the voltage gain and output and input resistances of the common- 
source and common-drain circuits. 

6.15 The common-source circuit is the counterpart of what bipolar transistor cir- 
cuit? Why? 

6.16 The common-drain circuit is the counterpart of what bipolar transistor circuit ? 
Why? 

6.17 An FET has# m = 6000 /zmhos at the Q point. In the common-drain circuit, 
what output resistance will be obtained ? 

PROBLEMS 

6.1 For the curves of Fig. 6.5(a), with V uu = 20 V, R = 4000 fi, and V os = -1.5 
V, find the Q-point current /„. 

6.2 Determine R s if the circuit of Problem 6.1 is to be self-biased. 



144 



The Field-Effect Transistor 



6.3 Design a self-bias circuit for a common-source amplifier using Ihe transistor 
in Fig. 6.9(b), with V DS = 12 V, I D = 3 mA at the Q point, and V DD 20 V. 
Determine R and /?.?. 

6.4 An FET, Fig. 6.9(b), is in a common-source circuit with Q point at V DS = 7.5 
V, I D = 4.0 mA, and V DD = 15 V. Find fl s , 7?, V os , /?,, and /{ 2 for R, + /? 2 
= 1.25 MQ. 

6.5 The 3N158 FET, with characteristics given in Fig. 6.7(a), is used in a common- 
source circuit with V DS = 10 V, /„ = 1.4 mA at the Q point. What should be 
the values of R, V D t» and V co for a good circuit design? 

6.6 Design a four-resistor bias circuit as in Fig. 6.10, using V DD = 30 V, V os = 
—2 V. Select I D and suitable values for the four circuit resistors. 

6.7 A depletion-mode FET has I DSS = 3 mA, V t = -4.25 V; find /„ for V os = 
—2.5 V by the square-law relation of Eq. 6.12. 

6.8 A depletion-mode FET obeys the square-law relation of Eq. 6. 1 2 with l DSS 
= 8.4 mA, V, = -3.0 V. Find I D at V os = - 1-0, -2.0, -3.0, and V and 
plot the transfer curve. 

6.9 Find g m at each V os value listed in Problem 6.8 by assuming A V os changes of 
0.2 V and computing the resultant A/„. Plot a curve of g m versus /„. 

6.10 What R should be specified for the circuit in Fig. 6.13(a) to obtain A v = -40 
with g„ - 6000 fimhos and r d = 40,000 Q? What is the output resistance? 

6.11 Given V DD = 30 V, /? 2 = 0.1 MQ, and /?, = 2 MQ for the transistor in Fig. 
6.10, find R s to place Q at I D = 5 mA, V DS = 15 V, and V os = -2 V. 

6.12 Determine/? for a JFET self-biased amplifier circuit as in Fig. 6.7(b), for a volt- 
age gain of 25, with r d ^ 40 kSl, g m 0.005 mho. Also, select a suitable 
value for R . 

6.13 Find R, R s , and R t for the circuit in Fig. 6.9(a), with a Q point for the transis- 
tor at I D = 4.0 mA, V DS = 7.5 V. 

6.14 An FET with r d = 50 kfi, R s = 2 kfi, and g„ 2500 //mhos is used in 
a common-drain circuit. What is the voltage gain? What is the output 
resistance? 

6.15 Calculate the ac voltage gain of a common-source FET amplifier in the circuit 
in Fig. 6.7(b), with g m = 0.0045 mho, r d - 50 kQ, and R = 20 kQ. 

6.16 What value of transistor # m is needed in the circuit in Fig. 6.7(b) to provide 
a signal gain of -40 if r d = 60,000 Q and R is chosen as 20,000 Q and R 
is I MQ? 

6.17 A signal V, - 2 mV is applied to the FET amplifier in Fig. 6.7(b) with g m = 
2500 //mhos, r d --= 40 k£i, R - 10 kfl, and R - 1 MQ. Find the ac output 
voltage V u . 

6.18 What value of load resistor R s would match the output of a common-drain 
amplifier with g m = 0.0025 mho, r d = 40,000 Q ? Find the voltage gain with 
that resistor. 



7 

The Vacuum Tube 






The triode, or three-element form of vacuum tube, was the first electronic 
control device. It employs a heated electron-emitting cathode, a control grid 
of wire mesh, and an electron-collecting plate or anode, assembled in an 
evacuated glass, metal, or ceramic envelope. Additional grids are added to 
improve the electrostatic shielding between grid and anode, leading to the 
five-element pentode. 

Because of the need for heating power to induce emission of electrons, the 
large size and fragility, and the limited life, the vacuum tube has been 
superseded by the transistor at frequencies below 500 MHz and power levels 
below 200-W output. Many millions of tubes remain in service, however, so 
we give a brief discussion of tube and circuit theory here. 

7. 1 Circuit Notation for the Vacuum 

Tube 

As for the transistor, there is an established system of nomenclature for 
circuit variables employed with vacuum tubes. The symbols lack the for- 
mality of those for the transistor, however. 

Again, we use lowercase letters to designate instantaneous values of 
varying currents or voltages and capital letters to denote rms or dc values. 



145 



146 



The Vacuum Tube 



Subscripts b and p indicate anode circuit quantities and c and g indicate grid 
circuit variables. For example, 

i h = instantaneous anode current 

I h = quiescent value of anode current 

I p = rms value of the signal component of anode current 

v c = instantaneous grid-cathode voltage, also v, k 
V cc — grid circuit bias voltage 
V BB — anode circuit supply voltage 

Some tubes have more than one grid and numerical subscripts are used, 
with the grid nearest the cathode as number l. 



7.2 The Triode 

The drawing in Fig. 7.1(a) shows the construction of the internal elements of 
triodes used at audio and low radio frequencies. There is a cylindrical nickel 
anode A, a helical grid G, and a central electron-emitting cathode K. The 
tube in Fig. 7.1(b) is designed for high-power transmitting service with air 
cooling and that in Fig. 7.1(c) is for high-frequency radio reception and 
illustrates a planar form of cathode and grid. 

If the free electrons within a metal are given sufficient energy, they are 
able to overcome the surface binding forces and can be emitted into space. 
The releasing energy can be supplied by heat in thermionic emission, by 



Glass Envelope 



Anode 




5 cm 




I cm 



Ceramic 



(a) 



(b) 



Figure 7.1 (a) Low-frequency triode structure; (b) Eimac 8874 triode, 1000-W 
peak output up to 500 MHz; (c) small ceramic triode, usable to 450 MHz. 



The Triode 



147 



radiant energy in photoelectric emission, or by bombardment by atomic 
particles in secondary emission, or the electrons can be pulled out of the 
surface by the attractive force of a strong electric field, as in cold-cathode 

emission. 

For thermionic emission, the emitting cathode is most often a small 
nickel cylinder coated with a layer of barium or strontium oxides and heated 
to about 600 to 800°C (1000 to 1400°F). A low voltage is applied to a heater 
wire of tungsten to raise the nickel cylinder to emitting temperature. Such 
a cathode is said to be indirectly heated. Some tungsten wire filaments, 
impregnated with thorium, are used and operated at temperatures of 1600 
to 1700X (3000°F). Wire filaments are heated by a current through the wire 
and are said to be directly heated. 

The electrons from the heated cathode reach the anode only when the 




(b) 





+. 

lo U- 



\o- 




-o2 



-oi 



(C) 



lo- 



ta) 




-o2 



lo- 



-o2 



(d) 

Figure 7.2 (a) Triode circuit symbol and bias sources; (b) common-cathode 
circuit; (c) cathode follower circuit; (d) grounded-grid circuit. 



148 



The Vacuum Tuba 



anode is positive to the cathode and the triode conducts in only one 
direction. In transit the electrons pass through the openings in the wire grid. If 
the grid is negative to the cathode, electrons are repelled and the anode 
current is reduced. If the grid is made less negative, the anode current 
increases. It is seen that the negative potential between grid and cathode 
controls the electron flow and consequently the current through the tube. 

Practically no electrons can reach the negative grid and hence there is no 
grid current. As a result the triode has an infinite input resistance at fre- 
quencies through the audio range. Effects of the capacitances between the 
tube elements become important at frequencies of a megahertz or more and 
will be considered in Chapter 8. 

The circuit symbol for a triode and the applied bias potentials are shown 
in Fig. 7.2. Again we have three internal electrodes and wish to connect them 
to four terminals or two output ports. As a result of the choice of common 
electrode, we have three basic circuits. These are usually designated as the 
common-cathode circuit, the common-anode circuit or cathode follower, and 
the grounded-grid circuit with the grid common. All are diagrammed in Fig. 
7.2. 

Our conventional current is again defined as positive inward and shown 
as i p from anode to cathode in the figure. 



7.3 Triode Character/sties 

The output curve family for a typical triode is drawn in Fig. 7.3(a). The 
region of linear, equally spaced curves is bounded by V c — and by the 
maximum rated anode loss or dissipation to maintain a safe operating 
temperature. The loss limit is determined by 



P*=VJ b (W) 



(7.1) 



and a limiting hyperbola is drawn for P d — 0.6 W in the figure. 

Opposite changes of anode voltage and grid voltage are offsetting in 
effect on the anode current. The anode current is more sensitive to the grid 
voltage since the grid is closer to the cathode than is the anode. We measure 
-this sensitivity by the amplification factor n, defined as the ratio of anode 
voltage change to grid voltage change needed to keep i b constant. That is, 






(7.2) 



Triode Characteristics 



149 





- 2 i b (mA) 



The negative sign arises because the anode voltage and the grid voltage must 
change in opposite directions to maintain constant current. 
Values of n for triodes range from 3 to over 100. 



- 12 



(b) 
Figure 7.3 (a) Output curve family; (b) transfer curve family. 

The reciprocal of the slope triangle of Fig. 7.3(a) defines the plate resis- 

(7.3) 



lance r p as 



r =^1 



160 



The Vacuum Tube 



This is the resistance presented by the triode to ac signal currents, and 
typical values range from a few hundred to over 100,000 Si. 

The transfer curves, relating input voltage and output current, are shown in 
Fig. 7.3(b). From the slope triangle shown we can define the transconductance 
J?„as 

A: -i 

(mhos) 



g. 



5i£l... 



(7.4) 



Transconductance is the control factor for the output generator, as it was for 
the transistor. The magnitude ofg m for vacuum lubes will range from 1000 to 
over 40,000 //mhos. 

The product of the magnitudes is 



^ " Wr A/, 



b 



Aw, 



or 



gm r . 



(7.5) 



(7.6) 



This shows that the three coefficients are related. A low-// lube will have a low 
plate resistance and a high-// tube will have a high plate resistance. 



7.4 The Pentode 

In the triode we find a small capacitance C„ of 2 to 5 pF between grid and 
anode and a similar small capacitance between grid and cathode. At fre- 
quencies of I MHz or more in the common-cathode circuit, the reactance of 



'#*• 



Xc. = 



1 



2*/C„ 

becomes small enough so that it feeds back an appreciable current from the 
high signal voltage at the anode to the lower signal voltage at the grid. This 
feedback current causes instability of the circuit gain and possible oscillation. 
The use of the triode in the common-cathode circuit is therefore limited to the 
audio frequencies. 

This defect of the triode was overcome by adding two shielding grids 
between the control grid and the anode. Such a lube contains a cathode, a 
control grid G„ a screen grid C 2 , a suppressor grid C„ and the anode; the 
resultant five-element tube is called a pentode. The pentode is useful to 
frequencies of several hundred megahertz because C 2 and G 3 , as electrostatic 
shields, eliminate the capacitance between control grid and anode. A circuit 
symbol and a common-cathode circuit for a pentode are shown in Fig. 
7.4(a). 



Triode Characteristics 



151 



\{ °2 




(a) 



20 



15 



1 .0 











==n 






A 
















\\ 



































































































0.5 

1.0 

- 1.5 
-2 

3 

-4 

6 



100 



200 
(b) 



300 



400 



Figure 7.4 (a) Pentode in common-cathode circuit ; (b) output characteristics, 
6EA8 pentode, V , = 125 V. 

Grid G z is maintained at ground potential for signal frequencies by the 
large bypass capacitor C,. A positive voltage is placed on G 2 , the screen grid, 
and this accelerates the electrons to a high velocity. Thus G z is able to accel- 
erate the electrons with its positive dc voltage but still serves as a grounded 
screen for signal frequencies. The suppressor grid C 3 is connected to the 
cathode so as to serve as another electrostatic screen between grid and 
anode. The high velocity electrons that pass through the openings in G 2 coast 
through Gj and reach the positive anode. 

The parameters for the pentode, //, g m , and r„, remain as defined for the 
triode. Because of the low slope of the output curves, the plate resistance is 
found to approximate 1 MQ. The anode current is almost constant with 






162 



The Vacuum Tube 



The Triode Common-Cathode Amplifier 



163 



changes in anode voltage for a given grid voltage and the pentode behaves as 
a current generator. In general, it is similar in characteristics to the FET. 

7.5 The Equivalent Circuits 

In Chapter 3 we developed the idea of an h equivalent for any two-port 
circuit, and in Fig. 7.2 we drew the basic connections for a triode as a two- 
port element. Therefore, the h equivalent can represent the triode as in Fig. 
7.5. For the //-parameter circuit we had Eq. 3.15 and 3.16: 

Vi-kfi+KY* (7.7) 

Iz - hfl> + KV X (7.8) 

and we can rewrite these equations with vacuum-tube quantities as 

V, = h,I c + h,V 



i„ - g m V. k + ^ 



(7.9) 
(7.10) 



That is, r p = \/h as the plate resistance and we previously showed /;,/, = 
SmVv 



Triode 




Figure 7.5 (a) //-Parameter model from Chapter 4 ; (b) equivalent circuit for the 
ideal triode; (c) pentode equivalent circuil. 



We reasoned that a negative grid attracts no electrons and so I c — for 
the triode. At low frequencies there is no feedback of voltage from output to 
input and so h, = 0. The input circuit is open and Eq. 7.9 is meaningless. 

Equation 7.10 represents the output circuit of the triode, as in Fig. 7.5(b). 
We have a load current /„, dividing into current VJr f through the plate 
resistance, and a controlled current source g m V, k , with its current dependent 
on the grid-to-cathode ac voltage V, k . The circuit in Fig. 7.5(b) represents the 
triode for ac and is the current-source equivalent circuil for the triode, as a 
small-signal linear amplifier. 

A pentode equivalent circuit is derived by the same procedure used for the 
triode. The current taken when values of r p exceed 500,000 Q is negligible, 
compared to the current taken in usual loads, however, and r„ may be 
dropped from the circuit with negligible effect. We have the pentode equiva- 
lent circuit in Fig. 7.5(c), for small signals at audio and low radio frequencies. 

The form of the equivalent circuits for triode and pentode is identical to 
the models previously developed for the bipolar transistor and the FET. 



7.6 The Triode Common-Cathode 

Amplifier 

With the cathode as the common element we have the common-cathode 
triode amplifier in Fig. 7.6. We replace the triode with its equivalent circuit 
between G, A, and K in Fig. 7.6(b). Except for the notation, the circuit is 



M °2 







I o — It 



© 2 




(a) 
Figure 7.6 (a) Common-cathode triode amplifier; (b) equivalent circuit for (a). 



164 



The Vacuum Tube 




identical to those already analyzed for the bipolar transistor and the FET 
so we shall merely state the results of circuit analysis here. 

Voltage Gain 

The circuit yields 

(7.11) 

The negative sign indicates a 180° phase reversal between V, and V . 

Input Resistance 

The value of R t is usually made about 1 MQ and at audio frequencies the 
input of the triode represents an open circuit. 

Output Resistance 

We short-circuit V„ making V, = V ik = and the current source is then 
open (zero current). As a result, from the 2,2 terminals we see only 

K = r„ (7.12) 

as the output resistance of the amplifier. 

The common-cathode circuit is generally used for medium gain at audio 
frequencies. 

7.7 The Pentode Common-Cathode 

Amplifier 

The pentode common-cathode circuit in Fig. 7.7(a) was generally applied 
because of greater gain and stable operation at radio frequencies. To main- 
tain stability the reactance of C, should be only a few hundred ohms at the 
lowest operating frequency. Capacitors C, and C and resistors R, and R, are 
parts of the bias circuits and are dropped from the equivalent circuit for the 
ac signal. 

Voltage Gain 

The equivalent circuit is simple and the voltage gain can be written as 



4. = Y = s- R 



(7.13) 



Loads as high as 100,000 Q are used and gains of several hundred are 
obtained. 



The Cathode Follower 



166 



*V„ 





(a) 



(bl 



Figure 7.7 (a) Common-cathode pentode amplifier ; (b) equivalent circuit for (a). 
Input Resistance 

With R, E 1 Mil, the input resistance can be assumed to be an open 
circuit. 

Output Resistance 

Following the method used for the triode, we have 



/?„ = /•„ 



(7.14) 



and this is very large. 

The pentode common-cathode amplifier is generally used up to fre- 
quencies of many megahertz. A disadvantage of the circuit was the need for 
C, and R, in bypassing the screen grid to the cathode. Grid biasing may be 
done by any of the methods employed with FETs. 



7.8 The Cathode Follower 

In the common-anode circuit or cathode follower in Fig. 7.8, we place the 
anode at ground for the signal but maintain it positive for acceleration of the 
electrons in the tube. Resistor R, is large and dropped from the equivalent 
circuit. Circuit performance is found to be similar to that obtained from the 
emitter follower or the source follower. Resistor R, may approximate 1 Mil. 



156 



The Vacuum Tube 



l°— K- 



I o- 




-02 



lo- 



K* 



V„± »«• 



C fc 



*»"rfO 



-o2 



lo- 



-I 2 



(a) 



(b) 



Figure 7.8 (a) Cathode follower circuit; (b) equivalent circuit for (a). 



Voltage Gain 

From the output circuit we can obtain 

' * m 'V, + R k 
Around the input loop 

and 

Substituting for V tk and rearranging, 

Clearing the denominator and dividing by r p , we have 

But usually /? t < /■„ and /?*/r p can be dropped as small compared to unity. 
Then we can determine the gain as 



A - V °- *■** 
' V,~\ + gn,R„ 



(7.15) 



Since g„R k > 1, the gain is less than but near unity. One side of the 
output signal is at ground potential and this may be an advantage. 

Output Res/stance 

The output resistance at the 2,2 port is low and approximates 



The input resistance is very high. 



R, = j- (O 



(7.16) 



The Grounded -Giid Amplifier 



167 



The cathode follower acts as an impedance transformer, from a high 
input resistance to a low output resistance, as did the emitter follower and the 
source follower circuits. 

The circuit is usually designed with triodes since with only unity gain the 
high fi of the pentode provides no advantage and the pentode requires an 
expensive screen-cathode bypass capacitor. Grid bias is obtained as with the 
FET. 

Example: Find the performance of a cathode follower using a triode with 
H = 30 and r„ - 14,000 Q, R k - 3000 £1. 
We have 

^=£ = Wo = °- 002,4mho 

With Eq. 7.15 we need 

g m R k = 2.14 x 10" 3 x 3 x 10 3 = 6.42 
Therefore 

6.42 



A,= 



= 0.87 



1 + 6.42 
as the voltage gain. The output resistance is 



7.9 The Grounded-Grid Amplifier 

The circuit in Fig. 7.9 is the groimded-gricl amplifier, similar to the C-B 
transistor amplifier. Grounding of the grid provides an electrostatic shield 
and reduces the transfer of energy between output and input circuits. The 
frequency range of the triode is extended to many megahertz by this circuit 
but the input resistance is low. 

Voltage Gain 



The voltage gain is 



A ' V = 



g„R 



l + gji. I ± 



and this is relatively low. We have assumed 1 + M = M- 

Output Resistance 

The internal resistance factors of Eq. 7.17 lead to 

Ro = r,(\ + g m R.) 



(7.17) 



(7.18) 




158 



The Vacuum Tube 





R 5 K. 



(a) 



(b) 
Figure 7.9 The grounded-grid amplifier. 



Input Resistance 

The input resistance is conveniently written as 

and this is a small resistance. 

The circuit provides an impedance transformer of low gain, operating 
from a low input resistance to a high output resistance. Again, a pentode 
offers no gain advantage; when used the grids are connected together giving 
the characteristics of a high-// triode. 

Example: With g m -- 2500 //mhos, r p = 18,000 £2, R, = 500 £2 and R = 
4000 Q, find the gain and resistances when this triode is used in a G-G 
circuit. 

By Eq. 7.17, 

2.5 X 10" 3 x 4 x 10' 



A.= 



1 + 2.5 X 10-' x 500 + (4 x 10 3 )/(18 x 10 3 ) 
10 



2.47 



-4.05 



By Eq. 7.18, 



By Eq. 7.19, 



R„ = 18 x 10 3 (1 + 2.5 x 10" 3 x 500) 
= 40,500 a 

R ' /■ i 4000 \ 

' 2.5 x 10 J V ' 18,000,/ 

= 489 ft 




The Cathode-Roy Tube 



159 



These values confirm the statements made concerning the relative magnitudes 
of the input and output resistances. 

7. 10 The Cathode-Ray Tube 

The cathode-ray lube is a vacuum-tube device used for television viewing and 
as a laboratory instrument for visualization of electrical voltages and cur- 
rents in circuits. It is built in an evacuated glass envelope and employs 
electron-beam deflection by electric or magnetic fields that vary with the 
signals applied. The tube includes an electron-emitting cathode and beam- 
focusing electrodes in an assembly called an electron gun, followed by two 
pairs of mutually perpendicular deflecting plates and a fluorescent viewing 
screen. When magnetic field deflection of the beam is used, one set of 
deflecting plates is replaced with a pair of coils, producing a magnetic field. 
A visible spot of light is produced at the point of impact of the beam on the 
screen; the color of the light is dependent on the fluorescent material with 
which the screen is coated on its interior face. A cathode-ray tube is dia- 
grammed in Fig. 7.10(a) and (b). 

The accelerating potential V a increases the beam velocity and brightens 
the spot of light. The velocity of the electrons is very high; upon impact, 



Electron Gun 

A, 



Deflection 
Plates 




- P 




Screen 



t 
D 



Screen 



<b> 



Figure 7.10 (a) A cathode-ray tube; (b) deflection system. 



160 



The Vacuum Tube 



part of the electron energy is converted to visible light and part to heat on the 
screen. A stationary spot can cause a burn on the screen. 

The geometry of the deflecting plate system in Fig. 7.10(b) can be used to 
determine the deflection D on the screen as 






(m) 



(7.20) 



With two pairs of plates at right angles, the position of the spot can be 
moved in both x and y axes on the screen. More usually a sweep voltage, 
which increases linearly with time, is applied to the .v deflection plates. A 
voltage applied to the y deflection plates will then appear as if plotted 
against time. The complete formation of a sine wave on the screen is shown in 
Fig. 7.1 1, for one sweep. By use of a sawtooth form of sweep voltage, as in 
Fig. 7.12, the figure can be repeated. 



Screen 



Sine 
Signal 



on | 2 
y Plates 



Figure 7.11 Plotting of a sine wave against a linear sweep voltage. 

In television, one set of deflection plates is replaced with a pair of coils 
producing a magnetic flux density B, perpendicular to the electron beam. The 
deflection is that of a magnetic field on a current and is 



£L- 



(m) 



(7.21) 



j_Ll m B 

For a given deflection we can use higher acceleration voltages and obtain 
brighter spots with magnetic deflection. 

Materials used for the fluorescent screen have varying properties of light 
persistence after bombardment by the electrons. We have screens with image 







Review Questions 



161 



persislence in microseconds to screens with image persistence measured in 
minutes. Various colors are also obtainable, with green or blue common in 
laboratory equipment, and the three basic colors available for color television 
screens. 




Figure 7.12 A sweep waveform. 

7.11 Comments 

Because of its broad frequency range and high gain, the pentode was once 
widely employed as a voltage amplifier. Triode use was confined to audio 
frequencies or for impedance transformation when used as a cathode fol- 
lower. The grounded-grid circuit still finds some application in high-power 
transmitting equipment. 

Transistors are now generally used, however, because of the advantages 
previously listed. 

REVIEW QUESTIONS 

7.1 Explain what is meant by V BB , V C a V^ l b , /„ i b , V, k . 

7.2 Explain how the grid controls the passage of electrons through a triode. 

7.3 What happens if a vacuum tube is operated with a positive grid? 

7.4 What amount of power is dissipated in the triode in Fig. 7.4(b) with a Q point 
at V c = -1.8 V, V b = 150 V? (Interpolate the curves.) 

7.5 If the maximum allowable P d 2.5 W for a triode, would a Q point at V b = 
250 V, V e -- -I V be desirable for the triode in Fig. 7.4(b)? 

7.6 What boundaries would you select for the placement of a Q point on the 
curves in Fig. 7.3(a) for distortionless amplification? 

7.7 Define ft, g„, and r p . How are they related? 

7.8 Triode A has a wide-spaced grid, tube B has a close-spaced grid. Which has 
the highest /*? Which has the highest r p ? Could you determine £ m ? 

7.9 What can you say about current amplification in a triode? 

7.10 What happens to the anode current if a triode is operated with a grid bias more 
negative than — V BB ln"> 

7.11 Why do we add extra grids to form the pentode? 









162 



The Vacuum Tube 



7.12 Why is a penlode said to be a current source? 

7.13 How do we accelerate the electrons through the screen grid? 

7.14 Why do we use bypass capacitor C, with the pentode? 

7.15 How large should C, be? Why is the value of C, determined at the lowest fre- 
quency of interest? 

7.16 Why would you select a cathode follower in place of a common-cathode cir- 
cuit? 

7.17 List the similar types of amplifiers, using bipolar transistors, FETs, and vac- 
uum tubes. 

7.18 Name two reasons for use of a grounded-grid amplifier. 

7.19 What condition must be assumed to justify -g m R as the gain of a pentode 
common-cathode amplifier? 

7.20 Why does the grounded-grid circuit operate with stable gain at many mega- 
hertz? 

PROBLEMS 

7.1 For the iriode in Fig. 7.3, plot curves of anode voltage on the ordinate and grid 
voltage on the abscissa, for several values of constant anode current. Measure 
AvJAv c ; what is this parameter called? 

7.2 The triode in Fig. 7.3(b) is biased to a Q point at l b = 1.5 mA, K c = -4 V. 
What is the peak value of a sine-wave signal that can be used without driving 
the grid into the positive region? In terms of operating within a small region of 
the curves, would this sine voltage really be a small signal? 

7.3 What is the // of the Iriode represented in Fig.7.3(a) in the region near v b = 
250 V, i b = 1.5 mA? 

7.4 The transconductance of a triode is 4000 //mhos and ft is 8. The grid voltage is 
kept constant; find the increase in anode current when the anode voltage is 
changed from 200 to 235 V. 

7.5 An equation for the output curves of a triode is 

i„ = K(Sv t + v b ) (mA) 

At »•» = 200 V and v, - - 12 V, the current is 6 mA. 

(a) What is the current at v„ = 300 V and v c = —8 V? 

(b) What v c is needed to return i b to 6 mA at v b 300 V ? 

7.6 Taking A changes in voltages, determine fi,g m , and r„ for the triode described 
by 

Amp, i b - 38 x 10-*(f 4 I 8i' c ) 

near v b = 200 V, v c = - 12 V. 

7.7 The triode of Problem 7.6 has an anode current which is to increase 0.3 mA 
when the grid voltage changes from -2 to -3 V; what change in anode volt- 
age must be made simultaneously? 



Problems 
7.8 From the following data taken on a triode, find // and g m : 



163 



i» (mA) 


«c 


v b 


8.2 





165 


6.0 





130 


6.0 


-1 


165 


4.1 


-1 


130 


4.1 


-2 


165 



7.9 A pentode has r p = 650,000 ft, g m 2500 /zmhos. What is the value of //? 
What load R should be used to obtain a voltage gain of 120 in a common- 
cathode circuit? 

7.10 For the pentode in Fig. 7.4(b), determine graphically the value of r„ for v c = 
-1.5 V and the value of g m near v b = 200 V, i? c = -1.5 V. What is the value 
of/*? 

7.11 For a certain triode with g m = 3300 //mhos, r„ = 5100 £2. 

(a) Find the anode current change produced by variation of the grid-cathode 
voltage from -2 to -6 V, at v b = 140 V. 

(b) What change in anode voltage will bring the anode current back to its 
original value, with v c = —6 V? 

7.12 A common-cathode circuit uses a triode with r p = 4000 £2, n — 8, and input 
signal V, = 2 V rms, ac. Find the output rms voltage across a 5OO0-£2 load. 

7.13 Starting with Eq. 7.15, derive a gain expression for the cathode follower in 
the form 



r P + flR k 

7.14 In the cathode follower in Fig. 7.8, R k = 1000 SI, g m = 0.003 mho, and fi = 
25. If Vi = I V rms, find the output rms voltage. Also find the output resis- 
tance. 

7.15 A pentode has g m -- 0.0035 mho, r p = 650,000 Q and is used in a common- 
cathode amplifier to provide 35-V output when the input signal V, = 0.05 V 
rms. What value of R is being used? 

7.16 A grounded-grid amplifier uses a triode having fi ■ -- 70, r p = 40,000 CI. With 
R, = 300 fi, R - 10,000 £2, find the voltage again and the input and output 
resistances. 



8 

Frequency Response 
of RC Amplifiers 



We have been assuming that the series capacitors in amplifiers are infinite 
and the shunt capacitances of transistors and tubes are ideally zero. A prac- 
tical amplifier uses nonideal capacitor values, however, and the effect on 
the operation is dependent on the reactance of the capacitors that varies 
with frequency. When we use multistage cascaded systems for greater overall 
gain, the frequency effects are compounded. 

The frequency response of amplifiers will be studied here in terms of 
the very commonly applied resistance-capacitance (RC) amplifier. 



8. 1 Cascaded Amplifiers 

Figure 8.1 shows a typical RC transistor amplifier of two stages, from A, 
to A 2 and from A 2 to A 3 . These stages may be analyzed separately, with the 
overall gain determined as the product of the individual stage gains. 
Output V 3i is 

K« = -A,.,V 22 (8.1) 

and 

V2i=-A,y u (8.2) 

Then by substitution for K 22 , we have 

y 3 2 = (-A„K-A„)y u 






The Amplifier Passband 



165 




Figure 8.1 Two-stage RC amplifier. 



and the overall gain between port 1,1 and port 3,3 is 

A — ^» a A 

'II 



(8.3) 



and in general 

^.(ov) = A„A„A„ . . . (8.4) 

We have previously shown that when the individual stage gains are ex- 
pressed in decibels, then the overall gain is obtained as the sum of the in- 
dividual decibel gains: 

dB, /<„,„.> = A vltn + A„ 1B + A, UB + •■■ (8.5) 



8.2 The Amplifier Passband 

We usually wish the output waveform to be the same as the waveform at 
the input to the amplifier and that the distortion be negligible. However, 
our amplifier input signal may consist of many frequencies distributed over 
a wide frequency band. For instance, a l-fis pulse signal contains frequency 
components of which those to 4 MHz arc important if we are to obtain an 
accurate reproduction of the input waveform. To have an undistortcd 
waveform, we must amplify all frequencies equally. If this is not done, we 
have frequency distortion in the amplifier. Amplification of the pulse to 
I MHz will produce a distorted, but recognizable, pulse form. 

If we plot the performance of an amplifier as a gain versus frequency 
curve, ideally an amplifier should have a curve that appears as a horizontal 



164 



168 



Low 



Frequency Response of ftC Amplifiers 
High 



0.707 A 



■ulroid) 
V (mid) 







*angc 


J i Bandwidth h 


Range 


Mid Frequency 




^ "V 




/ 


i i i l 


\ 



10 I00 I000 I0 4 10 s I0« /(Log Scale) 

(a) 



'(> (mid) 




-0.707/1 



U "lii.ll 



10 



100 



103 

(b) 



I0 4 



105 



Figure 8.2 (a) Gain-frequency plot for an RC amplifier; (b) plot for a direct- 
coupled amplifier. 

straight line over the desired frequency range. More practically, we obtain a 
curve like that shown in Fig. 8.2(a). 

We have been assuming that blocking capacitor C c will have sufficient 
capacitance to be a negligible series reactance in the equivalent circuit. 
Such capacitors must have reasonable cost and size, however, and there will 
always be some range of frequency from zero upward in which the blocking 
capacitor is too small to meet the ideal requirement, and represents an 
appreciable reactance. This is the reason for the fall in gain in the low- 
frequency range in Fig. 8.2. 

Also, we have been assuming the maximum frequency of the signal to 
be low enough that the very small internal capacitance of a transistor or tube 
and the stray mounting and wiring capacitance represent such high reactances 
as to be considered an open circuit to ground across the input or load resis- 
tances. There is, however, always some frequency above which these react- 
ances have an appreciable effect in shunting the load. This accounts for the 
fall in gain in the high-frequency range in Fig. 8.2. 

Only in the mid-frequency range arc we able to satisfy the assumptions 
on both series and shunt reactances and consider them negligible in effect. 
In this mid-range of frequency we obtain the predicted gain figures of the 
preceding chapters. 

The low-frequency range and the high-frequency range are almost 
always well separated in frequency and quite distinct, as in Fig. 8.2. We are 
able to analyze the amplifier response in each region independently and to 
determine and evaluate the circuit factors responsible for the fall in gain. 



The Frequency Plot 



167 




We arbitrarily bound the mid-frequency region of uniform gain by choos- 
ing limit frequencies,/, and/,, at which the gain has fallen to 0.707 or \/«J~T 
of its value in the mid-frequency region. These frequencies are called the 
cutoff, band-limit, or half-power frequencies. At/, and/ 2 the output power 
is one-half of the mid-range output. That is, 



and 

At/, or/j, 
and the power is 



V. = A, 

P -VI . 

*mld — -V? - 



(mid) V l 



R 



R 



(8.6) 



V'„ = 0.707/4 



■■(mid) 



P _ (V'o) 2 _ [0.707^ B(miJ ,K,]' _ 0.5M. <aM> KJ» 



(8.7) 







We see that 

Pt*-*5r*m ( 8 - 8 > 

as predicted for the band-limit frequencies. 

The amplifier passband is defined as the mid-frequency region, with a 
bandwidth given by 

bandwidth (BW) =/ a — /, (Hz) (8.9) 

A direct-coupled (DC) amplifier omits C c and its low-frequency response 
extends to zero frequency. It still has a high-frequency region due to the 
inherent capacitances of the transistor or tube. The gain-frequency curve 
appears in Fig. 8.2(b). 

In deciding to neglect one resistance or reactance as large or small with 
respect to another, sufficient accuracy is usually obtained if there is a ratio 
of 10:1 in the respective impedance magnitudes. For example, a resistor of 
1000 Q is in parallel with another of 10,000 fi. The combined resistance is 

R 10 3 X 10 4 _ IP 4 

" L x 10 3 |- 10 x 10 3 II X 10 3 



= 9I0Q 



If we drop the 10,000-Q resistor from consideration as large with respect to 
the 1000-Q resistor, we are saying that we have a resistance of 1000 £2 in- 
stead of the actual 910 Q. This error is less than 10 per cent and can usually 
be overlooked because of the larger circuit variations introduced by the 
parameters of the transistors and tubes. 



8.3 The Frequency Plot 

The frequency axis of a gain-frequency curve is usually plotted on a loga- 
rithmic scale. In this way each multiple of 10: 1 in frequency is given equal 
distance on the abscissa and a very large frequency range can be covered, 



168 



Frequency Response of RC Amplifiers 




BuiulwKlth 




J_ 



J. 



I0 5 



I0 ft 



I0 3 \Q* 

Frequency (Hz) 

Figure 8.3 A gain plot in decibels. 

as shown in Fig. 8.3. Each multiple of 10: l in frequency is called & frequency- 
decade. 

The mid-frequency gain does not vary with frequency and can be used 
as a standard. By comparing the low- and high-frequency gains to the mid- 
range gain, we can obtain a general gain-frequency curve, said to be nor- 
malized on the mid-frequency gain. To normalize a gain figure, we divide by 
the mid-frequency gain, to give terms such as 

■^i'(lo) . A,i M \ 



*r(mMl 



Mrnld) 



The value of these ratios in the mid-range is unity for all amplifiers and so 
the gains are normalized to l. 

The normalized gain figures are ratios and can be converted to decibels, 
as 



-, dB-20log 



(8.10) 



•"►(mid) | ^»i(mlJ> 

In the mid-frequency range the ratio is 1 and the logarithm of 1 is so that 
when stated in decibels the normalized mid-frequency gain is OdB. At the 
limit frequencies we have 20 log (0.707) = -20 log (1.414) = -3dB and 
so the limit frequencies are correctly called the — 3-dB frequencies. A 
normalized gain-frequency plot in decibels is drawn in Fig. 8.3. 

The phase angle of A ft „ m for a single C-E stage is 180°. Normalizing of 
A ctM and A v(bl) against A, imli) will result in their phase angles being normal- 
ized against 180° as a reference angle. The total phase shift will then be 
180° -r- 0,„ and 180° - hl . At low frequencies the total phase shift tends 
toward 180° + 90° - 270° at zero frequency. In the high-frequency range, 
the total phase shift tends toward 180° — 90° = 90° at a very high frequency. 

8.4 Low-Frequency Response 

The most commonly used amplifier is the /?C-coupled, common-emitter 
circuit of which Fig. 8.1 is an example. The purpose of each of the circuit 
elements should be understood since the four-resistor bias network is em- 






Low-Frequency Response 



169 



ployed for each transistor. As drawn, the circuit is too complicated to be 
readily analyzed and we break it down by stages for better understanding. 
For this purpose we consider stage 1 as between A, and A* and redraw that 
part of the circuit in Fig. 8.4(a). 

We assume that C E is large and represents a negligibly small reactance so 
that we can drop R E and C B from the equivalent circuit. The effect of an 
unbypassed R E will be considered in Sec. 8.7. We replace the transistor Q x 
with its equivalent circuit, the input resistance h,„ and the current generator 

fcKfc. 

We previously showed that the effect of R, and R 2 could be replaced by 
a single resistance R B , where 

R B = *'*» 



R, H R 2 

and R B appears in the circuit in Fig. 8.4(b). By proper bias circuit design, 
however, we can choose R, and R 2 so that R B ^> R h where R,, is the input 
resistance of the transistor, usually /»,,. The equivalent circuit in Fig. 8.4(c) 
follows after dropping R B . 




CO 

Figure 8.4 (a) One stage of the RC amplifier; (b) partially simplified; (c) 
equivalent circuit at low frequency. 



170 



Frequency Response of RC Amplifiers 



Wc first write the mid-range gain by assuming the reactance of Q. neg- 
ligible compared to /?,,; this is really the definition of a mid-range frequency, 
with capacitor C c represented as a short circuit. Then 



Vzz = -g m V»- 



RR 



ii_ 



(8.U) 



(8.12) 



•R + R„ 

Then the mid-range gain for the first stage is 

a V 22 _ a A A,. 

" mW) ~ V bt - g "R + R„ 

We now reinsert C c and derive the low-frequency gain for the circuit in 
Fig. 8.4(c). The output K 22 is 

V^^ = ~IiR(, 
and we need to find / 2 as a portion of the transistor current g„V hc . 




Figure 8.5 The current divider example. 

This question can be resolved by use of the current-division factor. In 
Fig. 8.5 the voltage across the parallel branches is equal and so we can write 
the equations 

RJl* 



V=R b l, - 



R„ I R„ 
In Eq. 8.13 we cancel R„ on each side and the branch current is 



I\ = 

Likewise, with Eq. 8.14 we obtain 



R. + R» 



R. 



/ 



(8.13) 
(8.14) 

(8.15) 

(8.16) 



R„ + R> 

The current /, divides out of / in proportion to R b , the resistance of the other 
path divided by the sum of the resistances R, + R b . 

Similarly, / 2 divides out of / in proportion to /?„, the resistance of the 
other path, divided by the sum of the two resistance paths R„ + R b . 

Equations 8.15 and 8.16 demonstrate the use of current-division factors. 









Low-Frequency Response 171 

Wc now write the reactance of C c as 

x =-L-= ' 

c caC c 2nfC c 
using co = Inf. The impedance of the A path is then 

Accordingly, with R as the resistance of the other path and R — /?,, — 
(jloiCc) as the impedance of both paths, we have current / 2 as a fraction 
ofg m V><: 



(8.17) 



(8.18) 



R + R„ ~ UlcoC c ) 



Dividing out R + R h , we have 
/. = g m Y>. 



r r i 



R + R, 



\-j 



1 



As a magnitude, this is 

h = g m v> 



The output voltage is 



coCrAR + RJ. 
1 



Vzz = -IiR„ = -g m V» 



V + lcoC c (R + R„)j 
I 1 



(8.19) 



(8.20) 



R 4 R, 



y l + IcoCrAR + /?i,) J 
Dividing out V b „ we find the low-frequency gain as 

1 



A V " - c RR » 

"rtlrt - y^ ~ 8m R ., ^ 



V 1 + LcoCciR + R„)\ 



(8.21) 
LcoC c {R -r R„) 

Comparison of the multiplier term with Eq. 8.12 shows that the multiplier 
is ^„( m , d ). Then the gain ratio is 

^•(lo) 1 

(8.22) 



*»<rald) 



(8.23) 



V 1 + IcoCcAR + R„)i 
There is an associated phase angle: 

» = aa - < cJ + RJ 

in addition to the 180° phase shift in A r(mU) . 

The effect of frequency on the gain is shown by the radical in the de- 
nominator of Eq. 8.22. If/= 0, co = 2nf = 0, and the second term in the 
radical becomes infinite, the gain is zero, with a phase angle of 180° — 90° 



172 



Frequency Response of RC Amplifiers 



= 270°. As frequency increases, the second term in the radical decreases 
with respect to I and the gain ratio increases toward the mid-frequency value 
of unity. 



8.5 The Low-Frequency Limit 

Let us examine the situation when the terms in the radical of Eq. 8.22 become 
equal; that is, at /, 

1 - „ ,„ ,*. . „ , (8.24) 



Then we have 



2nf x C c {R | R„) 



A, 



m. = 



I 



V2 



= 0.707 



(8.25) 



^Hmld) ^/l -f 1 

We see that the frequency defined in Eq. 8.24 is the lower limit frequency of 
the amplifier. Frequency/, is associated with the circuit elements chosen by 
the designer as 

fl = 2nC c (R + R„) (8 " 26) 

The lower limit frequency is dependent on the size of the blocking capacitor 
and the resistances through which it charges. These are factors that the 
designer can select to place the cutoff of the mid-frequency region where 
cost of parts and size considerations permit. 

We can substitute this value of©, = 2nf t into Eq. 8.22 and 8.23 to ob- 
tain a general expression 

•"»(io) L 



with a phase angle 



= taEr'4 



(8.27) 



(8.28) 



These expressions are important results. Any amplifiers having the same 
product of blocking capacitance and charging resistances will have the same 
frequency response. That is, a small capacitance associated with large 
resistances or a large capacitance with small resistance values can lead to 
the same limit frequency. 

A general curve for the low-frequency response of all RC amplifiers is 
plotted in Fig. 8.6, from the decibel values of Table 8.1, which is calculated 
from Eq. 8.27 and 8.28. To use the curve, one need merely know the/, 
frequency for a particular amplifier and the response at other frequencies 
is readily found. In order to show the gain increasing with frequency as a 
normal low-frequency response, the curve is plotted in terms of f/f,. Figure 
8.7 is the general phase response. 



The Low-Frequency Limit 



173 






dB 







Limit 














Frequency / 






Mid 


t 
Range 


^v (mid > 








/S 












/ 












20 




















Slope: 
- 20 dB/Decade 












30 


Jill 




I I 


MM 




I I 


I III 





0.05 0.I 0.2 0.5 I.O 2 5 10 20 

Figure 8.6 Low-frequency response in decibels. 



TABLE 8.1 Low-Frequency Response from Eqs. 8.27 and 8.28 









Phase Angle 


Total Phase 


/.// 


an 


dB 


(degrees) 


Angle (degrees) 


10 


0.1 


-20 


+ 84° 


264° 


5 


0.2 


-14 


79° 


259° 


2 


0.5 


-7 


63° 


244° 


1 


1.0 


-3 


45° 


225° 


0.5 


2 


-1 


27° 


207° 


0.2 


5 


-0.2 


IT 


191° 


0.1 


10 


-0.04 


6" 


186° 



"Including the inherent 180° phase shift of a C-E amplifier at mid-frequency. 

The table and curve show that for a 1:10 change in frequency (I decade), 
at low frequencies, the gain changes by 20 dB. It is hardly necessary to plot 
the curve below 0A/lf, because we know that Ihe gain varies at 20 dB per 
decade and 



W 



dB 






0.1 

0.01 
0.001 



-20 
-40 
-60 




174 



Frequency Response of RC Amplifiers 



100° 

80° 

60° 

40' 

20 c 
C 







































































1 1 


1 1 II 




1 


iTiT 





0-1 0.2 



0.5 



10 



20 



.0 2 5 

fff, 

Figure 8.7 Phase angle variation ai low frequencies. 

The expression of Eq. 8.27 can be plotted for any amplifier by drawing 
a horizontal asymptote at dB for frequencies above/=/, and a sloping 
asymptote falling 20 dB per decade below f = f,. At/ = /,, the limit fre- 
quency gain is plotted at — 3 dB. The response can then be sketched through 
the — 3-d B point to the asymptotes. 

Example: The first-stage circuit in Fig. 8.1, between A, and A 2 , has 
C c = 0.5 /xF, R -= 5 kQ, and a transistor with h lt »■ 900 Q. Since 

r> R i*z 10 X 10 3 X 10 X 10 } _ Q ,nnr» 

K " ~ rTTrI ~ no x io' yiuu " 

this resistance can be neglected with respect to h lt at 900 Q, and R,, = h„ 



'U(lo) 



A 



ri iiu.l i 



dB 



5 

10 

- 15 
20 

- 25 






Limit 


V 








Fi 


equency 












\ 

-dB Pom 








/ 












Slo 


)c: 








A i i i 


20 dB 


Decade 

i i 


i i i i 




1 L_ 



20 



100 



200 



500 



50 
/(Hz) 
Figure 8.8 Response plotted for the example. 



Low-Frequency Response for the F£T and Vacuum- Tube Amplifiers 
= 900 £1. Then 



175 



/,= 



I 



I 



2nCc(R + /?,,) 6.28 x 0.5 x 10 6 X (5000 | 900) 
1 



O.OTs? 



= 54 Hz 



This is the band-limit frequency. The result is plotted in Fig. 8.8, by use of the 
asymptotes and the — 3-dB point at/,. 



8.6 Low-Frequency Response for the FET 
and Vacuum-Tube Amplifiers 

In Fig. 8.9(a) we show an FET amplifier using a four-resistor bias circuit. 
We assume that C s is large and adequately bypasses R s so that the com- 
bination can be dropped from the equivalent circuit in Fig. 8.9(b). We again 
have 



R.= 



_ Mi 



Ri 



/?, 



Comparison of the equivalent circuit of the FET amplifier with that for the 
bipolar transistor in Fig. 8.4(c) shows the circuits identical, except for some 
of the resistor designations. Therefore the low-frequency analysis of Sec. 
8.4 and 8.5 applies to the FET. We need note that the limit frequency for the 
mid-frequency region of the FET amplifier is written 

I 



/.= 



2nC c (R + R B ) 



(8.29) 



As a rule R + R n is much larger than for the bipolar transistor and the 
needed value of C c will be reduced for the same/, frequency. 

Figure 8.10 shows a pentode amplifier with its equivalent circuit. Once 
more we compare the equivalent circuit with Fig. 8.4(c) and see that it is 
identical to that analyzed for the low-frequency response of the bipolar 
transistor. For the pentode the limit frequency will be written 



/. = 



1 
2nC c (R | R K ) 



(8.30) 



and all the universal gain and phase curves apply with/, as the limit fre- 
quency for the mid-range of the pentode amplifier. 

Example: Typical values for a pentode amplifier would be R — 50,000 £2, 
R„ - 500,000 $2, and g m = 0.0035 mho. Find the value needed for C c to 
make/, = 54 Hz as in the example of Sec. 8.5 and find the mid-frequency 
and/, gains. 




176 



We have 



Frequency Response of RC Amplifiers 



V DD 



Cr 



(~)*o 



W* 



m> 



«il v n 



i 

\2- 



Rts n 



■ 



(a) 










Figure 8.9 (a) FET amplifier; (b) equivalent circuit. 



/■ = 



1 



2nC c (R -| R,) 

and R \ R, = 50,000 + 500,000 = 0.55 x IO 6 SI. Then with /, - 54 Hz, 
we have 

C 1 

c ~ In X 54 x 0.55 X 10 6 

= 5.36 X 10-» = 0.00536 x IO" 6 F 

= 0.005 /zF 

This is much smaller than the value of C c = 0.5 jxF needed to give the same 
limit frequency for the bipolar transistor. The voltage rating required for the 
0.005 ftF capacitor in the pentode circuit may be 300 V and that for the bi- 
polar transistor may be 15 V, however, so that costs and size may not differ 
appreciably. 



The Unbypessed Emitter Resistor 



177 











(b) 

Figure 8.10 (a) A pentode amplifier; (b) equivalent circuit. 
The gain at mid-frequencies is 

0.0035 x 5 x 10* x 0.5 x 10 6 



0.55 x 10 6 



= 159 






The gain at /, = 0.707 X A,, mW = 0.707 X 159 = 112. 

8.7 The Unbypassed Emitter Resistor 

While we have assumed the emitter bypass capacitor C E to be of sufficient 
capacity to bypass R E adequately, the capacitor is not often used, for reasons 
of space and cost. Then we have the C-E amplifier input circuit as in Fig. 



178 







Frequency Response of RC Amplifiers 

'c 



The Unbypassed Emitter Resistor 



179 




(a) 



(b) 



Figure 8.11 (a) Unbypassed emitter resistor; (b) equivalent circuit. 
8.11(a). The input circuit equation is 

V, = /»„/* + (/ 4 + I C )R E (8.31) 

Since l e = A /c / & , we have the input resistance R, as 



R, = jf = K + (I + h,.)R E 
= *<< + hf,R E 



(8.32) 



The input circuit is drawn in Fig. 8. 1 1(b), as part of the usual transistor equi- 
valent circuit, and consists of/;,, in series with a large resistor h f ,R B . Actually, 
this is the input circuit of a C-C amplifier. 
We have 



and 



ht, + h [t R t 



V -hi "it* i 

V " ~ Kh ~ h,. + h,jt E 



v, 



v, 



(8.33) 



1 + 



&R, 



+ iJt* 



The load voltage is V = — g m RV bt and substitution of Eq. 8.33 for V h , gives 

1 



y. = -g m Ri 



■v, 



i + g m R B 

But the term -g m R is recognizable as A v{mU} and so with R E in the circuit 
we have a gain designated as A',: 



A' — l-£ — A 



1 



K, -».«-«*» l+ gm R B 



(8.34) 



The gain is reduced by the presence of R E . 



Similar expressions are obtained for the gain with an unbypassed source 
resistor R s for an FET or for an unbypassed cathode resistor R K for a tube, 
used in the circuits in Fig. 8.12. 

When these bias resistors are not bypassed, the gain expressions of the 
preceding sections are multiplied by the factor 1/(1 + g„R E ), or equivalent, 
derived above. Using appropriate g m and resistance values, this factor is 
the same for all three active devices. 

The gain reduction is often made up by additional gain elsewhere in 
transistor circuits since the cost and space requirements for the bypass 
capacitor are excessive. With the relatively expensive and bulky vacuum tube, 
the extra gain was more costly and the bypass capacitor was commonly 
employed. 



'DO 








-02 



l o- 




S 

R s 



I o- 



-°2 



t n 


a 









+ 


^ 


o°» 


v» 




■s 






5 




K< 


|o 


» 


s4 




—< 



(a) 



(b) 




oi 



o2 




(c) 



(d) 



Figure 8.12 (a) and (b) FET with unbypassed source resistance; (c) and (d) 
triodc with unbypassed cathode resistor. 






1 80 Frequency Response of RC Amplifiers 

8.8 High-Frequency Equivalent Circuits; 

the Miller Effect 

The internal shunt capacitances of transistor and tube become small react- 
ances that cannot be neglected at some high frequency. A modified equi- 
valent circuit for the three devices is the result. 

For the junction transistor, the effect of the capacitances is illustrated 
by use of the hybrid-n equivalent circuit shown in Fig. 8.13. Capacitance 
C b , holds the charge stored in the base and is of many picofarads. Capacitance 
C bc is that of the depletion region at the reverse-biased base-collector junc- 
tion and is only a few picofarads. Resistance r , — \jh ot is large and eliminated 
in the equivalent circuit as we have done before. 




Figure 8.13 Hybrid-n transistor circuit. 



Figure 8.13 shows the capacities in the input circuit. The currents add 
at *as 

/, + /» = /, 

where 



(8.35) 



h = 



* = <oc te (y -v bt ) 



But V„ = A v V b , and we substitute, giving 

U = °>C bc (A v V bt - V bt ) = coC be V bt (A,- 1 ) (8.36) 

Substitution of I, and /, into Eq. 8.35 yields 

/,=/,- I z = <oC b .V bt - o>C bc V bl (A v - 1) 
Since A v = —g m R for the C-E circuit, the terms can be combined to give 
/, = o>V bt [C b , + (1 + g m R)C be ) (8.37) 

Current /, passes into an apparent input capacitance 

Q. - C„. + (1 + g m R)C bc (8.38) 



High-Frequency Equivalent Circuits; the Miller Effect 




181 



This capacitance is the result of a multiplied value of C bc , moved into the 
input circuit in parallel with C bt , between B and E. The effective value of 
C„ is much larger than either C bc or C b , and this multiplying of capacitance 
is called the Miller effect. 

We then have the two branches of the input circuit in Fig. 8.14, which is 
the high-frequency g m model of the transistor. For a given l b the voltage V b , 
will fall as the reactance of C lr falls with frequency. Therefore, the input 
capacitance C lt is responsible for the reduction of gain of the transistor 
amplifier at high frequencies. 



B 

o- 



lc 






'I... 



9lC k 



C 

-o 



Qg m y„r 



E E 

Figure 8.14 The high-frequency g m model. 

The FET has a capacitance C„ from the gate to the source through the 
insulating layer of a MOSFET or the depletion region of a JFET. The FET 
also has a capacitance C, d between gate and drain which includes the capaci- 
tance to the mounting of the transistor. These capacitances are usually in 
the range of I to 5 pF. The input circuit of the FET is drawn in Fig. 8.15(a) 
and is seen to be identical to the branched capacitances of the junction tran- 
sistor in Fig. 8.13. The FET, of course, has no equivalent of A,,. 

We could perform an analysis similar to that of Eq. 8.35 to 8.38 and 
would find that the capacitances of the FET can be represented by a single 
shunt capacitance 

C„ - C„ + (1 + g m R)C, d (8.39) 

much larger than C, d alone. The Miller effect also appears with the FET 
and the high-frequency model for the FET is that shown in Fig. 8.15(b), 



■*d 



+ / 1 



D 

+ 



•R 



i 

i 
t 










D 


+ 






+ 





Cvi 






(a) 



(b) 



Figure 8.15 (a) Capacitances for the FET; (b) high-frequency FET model. 




182 



Frequency Response ol RC Amplifiers 



with C,f in shunt to the input terminals. At high frequencies the input im- 
pedance of the FET is not infinite but falls as frequency increases due to 
C lf . With a driving voltage source having an internal resistance, the current 
taken by C lf reduces the value of V„ with increasing frequency and the gain 
falls. 

We might carry through the same reasoning process with the triode or 
pentode in Fig. 8. 16. We find a Miller-effect capacitance across the grid- 
cathode terminals as 

C„ - C t * + (I + g m R)C„ (8.40) 

We have a high-frequency model for the vacuum tube as in Fig. 8.16(c). The 
capacitance C„ is responsible for the fall of gain with increasing frequency 
as for the FET. 

Again, our three active devices operate in an equivalent manner. 

Example: A junction transistor has C bc = 4 pF, C b , = 60 pF, and g m = 
4 mA/V = 0.004 mho and the amplifier load is 10,000 Q. Find the input 
capacitance C, c . 

First 

g n R = 0.004 x 10 4 = 40 

Using the relation of Eq. 8.38, 

C u = C b , | (1 + g m R)C ic 

C„ - 60 + (! -r 40) x 4 = 220 pF 



l >"->< 



-\ A 



G — l- 



>< 




c«* 



A 




C 

c— 



(„--■- QgnV* 



A 

-o 



K K 

(a) (b) ( C ) 

Figure 8.16 (a) Capacities in the triode; (b) circuit of capacitances; (c) high- 
frequency vacuum-tube model. 



8.9 High-Frequency Response 

In the high-frequency range of an amplifier the series blocking capacitors 
represent zero reactance; this is also true for C e bypasses for the emitter 
resistors. But the internal C, capacitances of the active devices appear in 
shunt and we draw the first stage, A, to A lt from Fig. 8.1 in the high- 



High-Frequency Response 



183 




'b { 




(a) 



(b) 




Figure 8.17 (a) Stage 1 of the amplifier of Fig. 8.1 ;(b) high-frequency equivalent 
circuit. 

frequency g m model in Fig. 8.17. Resistance R t , is the input resistance of 
Q lt normally expected to be h le , and C„ is the Miller-effect capacitance of Q z . 
The gain of the C-E amplifier is 

A v = -g m Z„ 

where Z„ is the parallel impedance of R, and R h and the reactance of C„. 
This parallel impedance has a magnitude 

RR,. 






Zpl J(R + R 



__RR Jl _ 



1 



R + R,. 



The voltage gain in the high-frequency range of the amplifier is 



<4« 



(hi) 



_ gmRR, 



which has a phase angle 



(8.41) 



(8.42) 



(8.43) 



The first term of Eq. 8.42 is the mid-frequency gain of the amplifier, 
^rtaid). however, and so 

■^r(hl) — -^i'(mia) — , . „ „,. . ? (8.44) 



The gain ratio follows as 



''ilnildl 









(8.45) 



184 



Frequency Response of RC Amplifiers 



The term in the parentheses in the radical increases with frequency /and the 
high-frequency gain falls. 

When the terms in the radical are equal, 
2nf 2 C lr RRi, _ | 
R + R„ 
we find the upper limit frequency f z as 



(Hz) 



(8.46) 



InCuRR,, 
R + R„ 

The upper band-limit frequency is dependent on the product of C u and the 
parallel value of the associated resistances. The frequency response of the 
amplifier is established as soon as the values of C„, R, and R,, are chosen 
by the circuit designer. 

Using/ 2 from Eq. 8.46, we can write the gain ratio in the more general 
form 

-^M- = ' (8.47) 

^m VI + (//A) 2 



with a phase angle 



* = tan-(-Z) 



(8.48) 



The frequency response in Fig. 8.18 is the general response curve for Eq. 
8.47 in decibels; the curve in Fig. 8.19 shows the phase response from Eq. 
8.48. Knowing the component values that produce a given f 2 frequency, 
the response curve can be plotted by use of a few points selected from the 
curve or values from Table 8.2. 



dB 





- 10 



Av 



''U0.it 

Imuli 



- 20 



-30 











1 

„ Limil 














S. Fre 


quency 














^ 


\ 
















A 
















Slope: \ 














20dB/de 


:aae ' 


\ 


! ' 


' 


1 ' 


1 1 1 1 


. L . 


1 | 


i i i i 


i 



0.05 0.1 0.2 0.5 1.0 

m 



10 20 



Figure 8.18 High-frequency response in decibels. 






The Frequency Limit of the Transistor 





20° 

"40° 

60° 
80° 
100° 



185 









































































: 


i i 


i i i i 


I 


i 






I i i i 


I... 



5 10 20 



0.I 0.2 0.5 l.O 2 

Figure 8.19 Phase angle variation at high frequencies. 

TABLE 8.2 High-Frequency Responses from Eqs. 8.47 and 8.48 






flh 



Decibels 
(gain) 



Phase Angle 
(degrees) 



Total Phase Angle* 
(degrees) 



0.1 


-0.04 


-6° 


174° 


0.2 


-0.2 


-11° 


169° 


0.5 


-1 


-27° 


153° 


I 


-3 


-45° 


135° 


2 


-7 


-63° 


117° 


5 


-14 


-79° 


101° 


10 


-20 


-84° 


96° 



•Including the inhcrenl 180° phase shift of a C-E amplifier at mid-frequency. 

The behavior of the amplifier at high frequencies is the inverse of the 
behavior at low frequencies when frequency ratios are used; that is, at 0.5/, 
the gain is the same as the gain at 2/ 2 , using the mid-frequency gain as a 
reference. 

At higher frequency ratios the gain continues to fall at the rate of —20 
dB per frequency decade. 



8. 10 The Frequency Limit of the 

Transistor 

Common-emitter current gain h fl falls at high frequencies, primarily because 
of the internal capacitances represented by C,,. If we use h ft0 as the usual 
gain figure at low or mid-frequencies, then h f , at any frequency is given by 






186 



the relation 



Frequency Response of RC Amolitiers 



*/.= 



yr 



1x^_ 



Wf,Y 



(8.49) 



When compared with Eq. 8.47, the form of this equation indicates that 
f=f„ is a limit frequency or -3-dB frequency for the transistor, at which 

By definition, h fl is the current gain for a short-circuit load, as in Fig. 
8.20. With R = 0, the Miller-effect equation is 

C„ = C h + (l+ fc x?)C fc 

= c 4 , + c„ l "^ 

This equation also comes from the circuit, Fig. 8.20, since C bf and C bc are 
placed in parallel by the short-circuit load. 







if 




/„ 


+ 




ii 






V »r i 


r* i 


'■Ct, 


C 


J «m "'ir 


L ° ' 











Shorl Circuit 



Figure 8.20 The transistor at high frequencies, under short-circuit load. 

With short-circuit load there are no circuit elements external to the tran- 
sistor to downgrade the performance. Therefore, with Eq. 8.50 as the least 
possible input capacitance, the short-circuit limit frequency for the transistor 
is defined as 

1 



/,= 



(8.51) 



2n(C b . -i C bc )h lt 

A transistor is operable at frequencies above/, but at values of A,, below 
h ft0 , as shown in the plot of h f , against///) in Fig. 8.21. The value of/, is 
used as a frequency figure of merit to compare transistors. 

With frequency/ well above/,, the value of///", > 1 and the denominator 
of Eq. 8.49 simplifies to give 

h, 



h " ~ m 



(8.52) 



The h [t curve at large//, values falls at -20 dB per frequency decade and 
if the curve is extended to a frequency / = f T , where h ft has fallen to unity 
or OdB, we have another frequency limit of the transistor, f T . Then from 
Eq. 8.52 



1 = h f'« 

TWe 

fr = ffh/t 



(8.53) 



The Frequency Limit of the Transistor 






187 






35 








J l 










30 








S 










J.J 










1 


.0 dB/De 


cade 




X 20 
pa 
















^ 15 


















10 














h\c-/-: 




5 

o 










c-h 




> 


% 


5 




. 1 1 


i i i i 




i i 


1 1 ! 





0.1 



0.2 



0.5 



10 



20 



50 



f/f. 



Figure 8.21 High-frequency performance of a transistor. 



Substituting/, from Eq. 8.51 gives 



fr = 



fl /fQ 



Q I 



(8.54) 



2n(C b . -|- Cjh, t 2n{C b , + C bc ) 

since g m0 = h t Jh„. 

The transistor bandwidth is/ z — /, but, for a transistor,/, is at zero fre- 
quency and fi = /, so that 

BW =/, (8.55) 

for the transistor. Equations 8.53 and 8.54 represent a product of gain and 
bandwidth. This gain-bandwidth product, f T , is dependent only on transistor 
parameters and is a constant for a given transistor. Values of/, or/ r are 
given by the manufacturers and both serve as figures of merit, useful in selec- 
tion of a transistor for a given frequency range. 

When connected in a practical circuit the frequency limit becomes /, 
because of the presence of a load R and the Miller-effect capacitance. The 
first step in selection of a transistor is to determine that it has a needed value 
of h f , and the second step is to determine that/, is well above the expected 
highest operating frequency. If f T is given in the transistor specifications, 
Eq. 8.53 may be used to find/,. 

We have a similar gain-bandwidth figure of merit for the FET: 



G-BW=/ r = 



Sm 



2n(C„ 4- C„) 
and for a triode or pentode vacuum tube: 



G-BW = 



,S >.; 



2n(C,„ + C„) 



(8.56) 



(8.57) 



188 



Frequency Response of RC Amplifiers 



Example: Consider a transistor with/,- rating given as 20 MHz. This 
gain-bandwidth product tells us that a current gain of 100 is theoretically 
possible with a bandwidth of 200 kHz or that a current gain of 10 is ob- 
tainable to a frequency of 2 MHz. Another transistor having f T = 5 MHz 
would have a possible current gain of 10 to a frequency of only 0.5 MHz. 



8.11 The Common-Base Connection 
at High Frequencies 

By use of the figures of merit, we can compare the frequency range of a 
transistor with the emitter common to one with the base common. At mid- 
frequency we have a mld as the current gain of the common-base configuration 
and at any frequency 

a = , a -"' (8.58) 

The frequency/, is identified as the upper band limit or — 3-dB frequency of 
the transistor alone, in the short-circuit common-base connection in Fig. 
8.22. 




fl = 



Figure 8.22 Transistor in the C-B circuit with a short-circuit load. 

The short circuit is across C tc and eliminates that capacitance from the 
circuit so that 

C„ - C b . (8.59) 

Also 



"lb — 



(8.60) 



1 + h f . ~ h ft 

and the input resistance is much less than that of the transistor with emitter 
common. The reactance of C b , remains large with respect to its shunting 
resistance to a higher frequency with base common. Using C„ and h lb , we 
have the limit frequency/, defined as 

1 h, 



/.= 



— 2a 



_ g„ 



(8.61) 



2nC,(hJh f ,) 2nCJ, lt 2%C bc 
Comparison of Eq. 8.61 and 8,52 shows that because of the reduction in 



Bandwidth of Cascaded Amplifiers 




189 



input capacitance 

/. > hfJf ( 8 - 62 > 

and by reference to Eq. 8.54 

/. >/r (8.63) 

Comparison curves for the current gains of a transistor connected with 
emitter common and with base common are shown in Fig. 8.2 1 . The common- 
base connection has a greater gain-bandwidth product than does the con- 
nection with emitter common. 

The parameter/, is a figure of merit for a transistor with base common to 
input and output. 

8. 12 Bandwidth of Cascaded Amplifiers 

The overall high-frequency gain for a number of stages of amplification may 
be written in decibels as 



Mhi) 



Mold) 



dB - 20 log {l ^ J lhyvll + 20 log n + {f ) fuyvil 



I 20 log . 



1 



+ 



[i + (flhcYV n 

v/hcref lt ,,f 2b ,f2 e , ... arc the limit frequencies of the respective stages. We 
invert the ratios, use a negative sign, and transfer the square root operation 
outside the logarithm: 






|,dB=-.O,og[l + (£) 2 ]-lO.o g [l + (0] 
-,O,og[, + (0]--. 



(8.64) 



A similar expression involving fiJf,fi b /f,fulf, ... can be written for the 
low-frequency response. The highest/, and the lowest/, frequency would 
primarily determine the limit frequencies of the complete amplifier. There- 
fore it does not pay to overdesign the separate stages. 

Consequently we often use amplifiers of n identical stages in cascade. We 
would have the gain ratio 



overall, 






(8.65) 



Vi -J- (flfz) 

for n stages, with/, the same for all stages. 

We define / 2 ' as the limit frequency of the overall amplifier; that is, at 
f=fi the overall gain ratio is \\sj~l ' . Then 



190 



Squaring and taking the reciprocal, 



Frequency Response of RC Amplifiers 



2- =. + (£)* 
& = V2"" - 1 

72 



(Hz) 



(8.67) 



where/, is the limit frequency of one stage. 

Similarly we derive an expression for/,', the low-frequency limit for the 
overall amplifier, as 



/■' = 



/, 



(Hz) 



(8.68) 



V2"" - 1 

where/, is the limit frequency for one stage. 

With gains plotted in decibels, the high- and low-frequency asymptotes 
for the gain curves have slopes of 20ft dB per decade and — 20n dB per de- 
cade, as shown for the high-frequency region in Fig. 8.23. 

Table 8.3 and Fig. 8.23 indicate the bandwidth narrowing that occurs 
as we cascade identical amplifier stages to obtain increased overall gain. 
For instance, with three stages the low-frequency cutolT or limit frequency 
is raised by a factor of almost 2 and the upper limit frequency is reduced 
to almost \. Thus by cascading three stages we have reduced the overall 
bandwidth to about 50 per cent of that of each stage. 

TABLE 8.3 Bandwidth Limits for n Identical Stages 

fufi overall limit frequency 
A.fi ■ limit frequencies per stage 



n 


/;//, 


n 


filh 


1 


i 


1 


1 


2 


1.56 


2 


0.65 


3 


1.96 


3 


0.51 


4 


2.30 


4 


0.43 


5 


2.59 


5 


0.39 



To restore bandwidth, it is necessary to overdesign each stage, in this 
example by doubling its/ 2 frequency and halving its/, frequency. 

If we have stages with differing values of/ 2 , as/,,/*, . . . , an approximate 
value for/ 2 ', the overall frequency limit of the amplifier, can be found by 

I 



s-WFF 



Frequency Emphasis and De- Emphasis 



191 










0.5 I 

(111 
Figure 8.23 Generalized n-stage gain curves. 

8. 13 Frequency Emphasis and De-Emphasis 

The variation of output with frequency obtained with RC circuits is applied 
in other ways, including audio system tone controls, pre-emphasis and de- 
emphasis circuits in FM transmitters and receivers, and in tape recording. 
In such applications the limit frequencies are usually called turnover fre- 
quencies. The several curves can be drawn by the asymptote techniques 
of Sec. 8.5 and 8.9. 

Figure 8.24 shows a circuit for bass boost in an audio amplifier. It operates 
with a gain of A and the circuit reduces the middle and high frequencies. By 



20dB/Decade 



■ WW 





(a) 



/"(Log Scale) 

(b) 
Figure 8.24 Bass-boost circuit. 







192 



Frequency Response of RC Amplifiers 



reducing these frequencies the circuit appears to boost the unchanged low 
frequencies. The attenuation begins at 

A = 2*(*, + Rz)C (8 ' 69) 

with/, usually chosen in the neighborhood of 1000 Hz. The transition ends 
at 



A = 



I 



2nR,C 



(8.70) 



The level of differentiation between low and high frequencies is the insertion 
loss, obtainable from Table 8.4 as a function of the respective turnover 
frequencies. 

The circuit in Fig. 8.25 illustrates a treble-boost circuit, which actually 
reduces the mid- and low-frequencies by the amount of the insertion loss. 
The effect appears to boost the level of the high frequencies with respect to 
the mid- and low-frequency regions. 



TABLE 8.4 Insertion Loss for Figs. 8.24 and 8.25 



hlfx 



Decibels 



ftlfi 



Decibels 



2 


-6 


20 


-26 


3 


-9.5 


30 


-29.5 


4 


-12 


40 


-32 


5 


-14 


50 


-34 


6 


-15.5 


60 


-35.5 


7 


-17 


70 


-37 


8 


-18 


80 


-38 


9 


-19 


90 


-39 


10 


-20 


100 


-40 





(a) 



ALog Scale) 

(b) 
Figure 8.25 Treblc-boosl circuit. 



Frequency Emphasis and De-Emphasis 



193 



The treble-boost insertion loss is found from Table 8.4 when the turnover 
frequencies are chosen as 



In 



R,R 2 



(8.71) 



h = 



+ R, 



litR.C 



(8.72) 



By using a variable resistor for R 2 of the bass-boost circuit, the value of 
/, can be continuously varied. The insertion loss varies and the bass boost 
appears to be changed. Similar action is obtained in the treble-boost circuit 
by variation of resistor /?,. 

In frequency modulation radio systems, the high frequencies of the signal 
are prc-emphasized by the circuit in Fig. 8.26(a), giving a result that is the 
inverse of the curve in Fig. 8.26(c). The values of R and Care determined from 
the time constant RC = 75 x 10" 6 s, resulting in a turnover frequency 
/, = \/RC = 1/(75 x 10") = 2123 Hz. At the receiver the de-emphasis 
circuit of Fig. 8.26(b) is used with the same turnover frequency, giving a fall 
at high frequencies as in Fig. 8.26(c). All signal frequencies above the turn- 
over are dropped by the amount of the pre-emphasis and their original level is 
restored; however, the noise frequencies originating in the transmission path 
and in the receiver are reduced by the amounts of the curve in Fig. 8.26(c). 
The signal comes out at its original level, while the noise is reduced in level. 

In tape recording the high frequencies do not record well and accordingly 
they are boosted in recording. At the playback the high frequencies are re- 
duced to normal using a circuit with the elements in Fig. 8.24. The resulting 
reproducer curve is drawn in Fig. 8.27. 



(a) 



I VWV- 

R 



r± K, 



dB 

S 

10 

15 

-20 

25 



















- 3dB- 


_5&i 


i/"=2i: 


3 Hz 












N. 




































i 


i<>i 




i ' 


i i 1 1 





0.1 0.2 



(b) 



0.5 1 .0 2 
/(kHz) 
(c) 



10 20 




Figure 8.26 (a) Transmitter pre-emphasis circuit, RC = 75 x 10"* s; (b) FM 
receiver dc-emphasis circuit; (c) dc-emphasis of the high frequencies at the 
receiver. 




194 



Frequency Response of RC Amplifiers 

































/|=S 


)Hz 




























































-/2 = 


1326 Hz 








i i . 


i.i .i.i. 




'■■ 



40 
35 
30 
25 
3 20 
15 
10 

5 



0.02 0.05 0.1 0.2 0.5 1 2 

/(kHz) 

Figure 8.27 Standard tape playback curve, 3.75 ips (9.5 cm/s). 

8.14 Review 

The frequency range covered by an /?C-coupled amplifier may be divided 
into three parts: the low-frequency region, the high-frequency region, and the 
frequencies between, called the mid-frequency region. In the last region, all 
reactances in the circuit arc negligible and the gain is constant with frequency. 
We use this region as a gain reference, or dB. If the gain is not constant 
across all signal frequencies, then we have an amplifier with frequency dis- 
tortion. 

In the low-frequency region the gain is affected by the scries blocking 
capacitor and the gain starts at a low value, rising at 20 dB per decade up to 
/, at which the gain is —3 dB. The frequency/, is determined by the blocking 
capacitance C c and the series value of the load It and h lr of the following 
transistor. 

The high-frequency region is bounded by a limit frequency f 2 at — 3 dB 
from the mid-frequency gain level; above/ 2 the gain falls, ultimately reaching 
a rate of — 20 dB per decade. The frequency / 2 is inversely proportional 
to C,„ the Miller capacitance, and to the parallel value of R and h lt . 

Each type of device has its own ultimate figure of merit as the gain-band- 
width product. This is a frequency at which the gain is unity, or OdB, and 
comparison of frequency performance is possible by using the G-BW 
figures of the devices. 



Review Questions 






196 



REVIEW QUESTIONS 






8.1 Why do we need to consider the frequency response of an amplifier? 

8.2 Define frequency distortion. 

8.3 Why are we able to consider the three frequency regions separately? 

8.4 What type of distortion is often present in RC amplifiers? 

8.5 For a rectangular 1 -fis pulse, repeated 60 times per second, there are harmonics 
every 60 Hz. How many harmonics must be reproduced by an amplifier for 
perfect reproduction of the pulse? For an approximate reproduction? 

8.6 What function does the blocking capacitor serve? 

8.7 The scries reactive clement determines what frequency limit? What frequency 
limit is fixed by the shunt reactances? 

8.8 What function does /?, and Ri serve in a transistor amplifier? 

8.9 What function does R c in an FET amplifier, and R„ in a vacuum-tube ampli- 
fier, serve? 

8.10 Draw the small-signal high-frequency model of an FET and label the elements. 

8.11 What is the effect if the emitter resistor R E is not bypassed? 

8.12 How would you choose the capacitance value of a blocking capacitor? 

8.13 What frequencies bound the mid-frequency region ? 

8.14 What factors determine/, for the low-frequency range? 

8.15 What factors determine / 2 for the high-frequency range? 

8.16 A transistor amplifier has/, 150 Hz; what is the gain at 300 Hz, with mid- 
frequency as the reference gain? 

8.17 A transistor has //„ - 800 £2; what maximum reactance may C c have in a 
well-designed amplifier at the lowest frequency used? 

8.18 In Question 8.17, why are we interested in the reactance of C c at the lowest 
frequency? 

8.19 An amplifier has a mid-frequency gain of 37 dB; its/, value is 15,000 Hz. What 
is the gain in decibels at a frequency of 45,000 Hz? 

8.20 What is the physical origin of the two internal capacitances in the high- 
frequency model of the junction transistor? 

8.21 Repeat Question 8.20 for the FET. 

8.22 Repeat Question 8.20 for the triode lube. 

8.23 What is the Miller effect? 

8.24 Is it a capacitance or a gain that is responsible for the Miller effect? 

8.25 Why do we use the short-circuit load to find the frequency limit of a transistor? 

8.26 Define/,; define/-; what is the relationship between/, and/-? 

8.27 Why does h, t of a transistor decrease at high frequencies? 



196 



Frequency Response of RC Amplifiers 



8.28 How should / 2 of an amplifier be related tof T of ihe transistor? 

8.29 Show that variation or R, in Fig. 8.25 will vary the high-frequency boost of 
an amplifier with a tone control. 

8.30 Why is a de-emphasis circuit used in an FM radio receiver? 



PROBLEMS 

8.1 In the circuit of Fig. 8.28(a) with transistors having /;„ = 2000 SI and h f , = 
40, find the value of C that will reduce the gain at 100 Hz by 3 dB below the 
mid-range region. 

8.2 An rtC-couplcd amplifier with a single stage has /", - 82 Hz,/ 2 = 12,500 Hz. 
Sketch the complete frequency-response curve, using a log scale on the fre- 
quency axis and gain in decibels. 

8.3 In the circuit of Fig. 8.28(b) the transistors have g„ = 0.015 mho and //„ = 
900 SI. What is the decibel mid-frequency gain from port 1,1 to port 2,2? What 
is/,? 



5 kft < I Mfi 



r€> 




Figure 8.28 

8.4 An FET with g m - 0.004 mho is used in the circuit of Fig. 8.29(a), with C„ = 
2 pF and C, d m 4 pF. What are the upper and lower half-power frequencies? 
What /, will produce 10 V at V in the mid-frequency range? 

8.5 A triode amplifier has C c - 0.005 fiF, R, = 0.25 MSI, r p = 27,700 £2, g m - 
0.0026 mho, and /? =- 0.10 MSI. Find A v{mU) ; also find A, at 100 Hz. What is 
/,? 

8.6 Using a transistor with h, e = 850 SI, h fc = 60, R = 5000 £2, C h , = 100 pF, 
and C bc = 3 pF, what is the transistor input capacitance? 

8.7 The transistors in the amplifier of Fig. 8.29(b) have h u = 1600 fi, h f , = 40, 
C be = 4 pF, and C„, = 100 pF. 

(a) Find/", and/ 2 for the amplifier. 

(b) What is the mid-frequency gain ? 



M 



Problems 



197 






lOkft 








Hf- 



4.7 kfi 




• 5.6 k« 



(a) 



(b) 



Figure 8.29 



8.8 A transistor (2, in the circuit of Fig. 8.30(a) has h„ = 1200 SI, g m - 0.005 
mho, and C = 0.05 (iF. Find /, and the gain at 30 Hz. 

8.9 The triode of Fig. 8.30(b) has r, 1 2,000 Q, g m = 2500 //mhos, C, k = 3 pF, 
and C pk = 3 pF. Find the bandwidth in hertz for the gain between ports 1,1 
and 2,2, with R = 40,000 Q for T x . 



60 kfl 




(a) 



(b) 



Figure 8.30 



8.10 Given the following transistor measurements at low frequency: 

//„ = 600 Si C bc = 3 pF 
A /r = 30 C bt = 100pF 

Find ff,f T , for the transistor. 

8.11 When the transistor of Problem 8.10 is used in the C-E amplifier of Fig. 8.30(a), 
find/ 2 . Also find the decibel gain at/= 2.5/i. 

8.12 Plot the high-frequency gain in decibels for Problem 8.1 1. Prove that your gain 
figure is correct at 2.5/ 2 . 

8.13 An amplifier has three identical stages with each having/, - 10 Hz and f 2 
= 25 kHz. What are the limit frequencies of the amplifier? 



198 



Frequency Response of RC Amplifiers 






8.14 Wilh V 80, C 6c = I pF, C*. = 75 pF, and /,„ . 600 £2, find f„ and f T or 
the transistor. What gain is possible with BW = 100 MHz? 

8.15 Specify the/, and/, design frequencies for each stage of a four-identical-slage 
amplifier if the overall bandwidth required is from 100 to 450,000 Hz. 

8.16 Show that a rate of fall of 20 dB per decade is equivalent to 6 dB per octave 
(double frequency). 

8.17 You wish the gain to fall at a rate of —60 dB per decade at a frequency that is 
high with respect to your operating frequency. How will you design your 
amplifier? 

8.18 An amplifier has/, = 80 Hz. Sketch the frequency-response curve on a log 
frequency scale from 10 to 500 Hz. What is the rate of gain fall at 0.1 Hz? 
Use a decibel gain scale. 

8.19 A transistor has/, = 10 MHz and h fco - 100 at a mid-frequency. Find It,, 
at 15 MHz. 

8.20 A transistor has/ r = 250 MHz and h fco - 64. Find h f < at 5.0 MHz. 

8.21 An amplifier has three nonidentical stages with gains of A t = 18 dB, A 2 *» 20 
dB, and A 3 = 14 dB at mid-frequency. What is the overall gain, expressed as 
An = VJV,1 

8.22 A two-nonidcntical stage amplifier has a first stage /, ^ 40 Hz and a second 
stage/, -= 15 Hz. Find/, for the overall amplifier. 

8.23 Plot the high-frequency region gain on a log frequency scale for a six-stage 
amplifier, with/ 2 = 1 MHz for each stage. Use a decibel gain scale. 

8.24 The G-BW product for a transistor is 120 MHz. With mid-frequency g m0 = 
0.010 mho and R - 2000 Q for the load, /;„ = 2200 fi. Determine the gain to 
be expected at a band limit of 4.5 MHz. 

8.25 A given amplifier has an/ 2 value of 30 kHz. At what frequency is the amplifier 
gain down only 0.1 dB from its mid-range value? 

8.26 For the amplifier of Problem 8.24, we find the gain is 1.5 dB below the mid- 
range gain. What is the frequency? 

8.27 A transistor has a mid-range a - 0.97 and/ = 0.55 MHz. Find the magni- 
tude of a at 0.75 MHz. 

8.28 A one-stage amplifier has A. = 30 and / 2 - 400 kHz. The /, value is to be 
increased to 600 kHz. What change can you make in the amplifier? What will 
the new A v become? 

8.29 An amplifier for an electronic voltmeter must have A, constant within 5 per 
cent up to 100,000 Hz. What/, value must the amplifier have? If composed of 
three identical stages, what/ 2 value must be specified for each stage? 






9 

Negative Feedback 
in Amplifiers 



In electronic circuit design we are often forced to sacrifice performance in one 
area to achieve better performance elsewhere; this is called a trade-off. In the 
use of negative feedback, we build extra gain into the original design and trade 
off this excess gain to obtain reduced distortion, stable gain, greater band- 
width, and changed amplifier input and output resistances. When we use 
solid-slate devices and integrated circuits, we find that the price paid for the 
additional gain is relatively low. Negative feedback makes our amplifiers 
nearly precise and ideal. 

9. 1 The Black Box with Feedback 

Negative feedback is an old and fundamental process, used in many fields to 
make the output response of a system more nearly correspond to the input 
signal. We compare the output with the input and utilize any difference as a 
corrective signal. The principle is apparent in such a simple activity as placing 
a pen on paper; without optical feedback of the difference between actual 
(output) and desired (input) locations of the pen, we could not write. The 
children's game of "pin the tail on the donkey" is another example of our 
inability to perform as usual without feedback in some form. 

In electronic amplifier feedback we compare a sample of the output wave- 
form against the input waveform. Any difference between the two signals 
gives an error voltage that is applied to the amplifier so as to change the output 



199 



200 



Negative Feedback in Amplifiers 



waveform and reduce the difference toward zero. Thus we force the output 
toward equality with the input and somewhat unprecise amplifiers become 
almost ideal gain elements that give an output like the input. 

The C-E amplifier in Fig. 9.1 has negative feedback applied through resis- 
tor R f and the blocking capacitor C f . A portion of V„ is transmitted back to 
the input circuit through R,. Because of the 180" phase inversion in the C-E 
amplifier, the voltage fed back as V f is opposite in phase or negative to V ( 
and the feedback voltage is compared with and subtracted from the input 
voltage. The difference becomes the amplifier input. 



Network 



Figure 9.1 A practical feedback amplifier. 
Figure 9.2 shows an amplifier system of gain 

(9.1) 



A V. 



from port 1 to port 3. The internal gain of the black box from port 2 to port 
3 is 

K 



-4 = 



v\ 



(9.2) 



The output is sampled by the /? network and the result V, is subtracted from 
the input at the mixing point £ (sigma), as shown by the indicated polarities. 



P ■* 



-±K 



Figure 9.2 Feedback applied to a black box amplifier. 



The Black Box with Feedback 



We have 



201 




(9.4) 



A K 



Vf - PK 

where /? is less than unity and the /? circuit is designed to produce zero phase 
shift in the operating frequency range. A resistance voltage divider serves well, 
as the R r , R b divider in Fig. 9.1. 
At the amplifier input, 

V, = V, - V, - V, - fiV (9.3) 

From Eq. 9.2 we have A V, =» V\ and substitution of Eq. 9.3 gives 

A(V, - pV e ) = V 

K;,(l -I Afi) - AV, 
We find the gain from port 1 to port 3 of the feedback system by use of Eq. 
9.1 as 

This is the fundamental equation of feedback, expressing the closed-loop 
gain A' as dependent on the internal gain A and on the feedback factor /?. 
In Eq. 9.3 the feedback voltage V f is presented to the input circuit in sub- 
tractive fashion. The denominator 1 1 + Afi\ > 1, and the feedback is nega- 
tive. Equation 9.5 then shows that | A' | < | A | and the gain of the system with 
feedback is less than the internal amplifier gain. Thus gain is sacrificed with 
negative feedback. 

If A is negative, as is usual in C-E amplifiers, we reverse V f from the /? 
network, resulting in a positive A/} term in Eq. 9.5 and so retain the negative 
feedback. 

If the phase of V f reverses, as may happen with nonresistive /? networks, 
the feedback voltage V f becomes additive to V, in Eq. 9.3 and the denomina- 
tor of Eq. 9.5 shows that 1 1 -\- Afl\< 1 and the feedback is positive. The 
closed-loop gain is | A' \ > \A\ and the gain of the feedback system is greater 
than the internal gain. This is a condition of gain instability since it includes 
the case at Afi = — 1 where the gain becomes infinite. The amplifier then 
becomes a generator of signals and when such action is wanted, we call the 
circuit an oscillator; these circuits will be studied in Chapter 13. The condi- 
tion of positive feedback is avoided in amplifiers. 

A measure of the amount of negative feedback introduced into an ampli- 
fier is given by the gain change in decibels as 



dB of feedback = 20 log -J 



(9.6) 






The latter expression can be translated into another that is useful. Look at 
? = ^j— = 1+j4/? (9.7) 



202 



Negative Feedback in Amplifiers 



so that the decibels of feedback can also be written as 

dB of feedback = 20 log (1 |- A0) (9.8) 

Because of the reduction of system gain by feedback, the input at port I, 1 
must be greater than the amplifier input at port 2, 2. By our definitions 

y. - A'V, 
V'o - AV\ 
From these relations, for equal output 



P.-yFj - £— K - (1 + Afi)V', 



(9.9) 



1 +AP 

The system input must be (1 I Ap) greater than the amplifier input at port 
2,2. 

Example: An amplifier has a gain A = 70 and a normal input signal of 
0. 1 V. Feedback with /? — 0. 1 is added, giving Ap = 7.0. The closed-loop gain 
is 

it A 70 _ one 

A "1 -V Ap~ 1 H-7.0~ 5 '° 
An input voltage 

K, = (1 f ^)K; = (1 + 7) x 0.1 = 0.8 V 
will be needed. 

The output voltage of the amplifier is 

V. = A'V, = 8.75 X 0.8 = 7.0 V 
or 

V' 9 = AV, = lQx 0.1 =7.0V 

These relations apply to the system and to the internal amplifier, at ports 
1, 1 and 2, 2, respectively. 

With feedback the amplifier input voltage is 

K ' :? - 70 ~° l V 

The external signal requirement is raised from 0.1 to 0.8 V by the addition of 
negative feedback, but inside the loop the amplifier is operating with an input 
of 0.1 V and an output of 7.0 V, with or without feedback. 

9.2 Stabilization of Gain by Negative 

Feedback 

When amplifiers are used in calibrated electronic instruments, such as volt- 
meters, internal gain changes will affect the instrument accuracy. Changes can 
be expected, due to supply voltage shifts, aging, and particularly operating 



Stabilization of Gain by Negative Feedback 



203 



temperature. Negative feedback is used to make the amplifier gain indepen- 
dent of these variables. 

For a percentage change in A, AA/A, we find a resultant percentage change 
in the system gain as AA'/A'. That is, 

Thus the feedback gain change is much less than the change in the internal 
amplifier gain. 

To reduce the sensitivity to internal gain change still further, we can make 
AP very large so that Eq. 9.5 reduces to 

1 



A'S 



P 



(9.11) 



With Ap ^> 1, we no longer are concerned with maintenance of an exact 
value of A because the overall system gain is dependent only on the elements 
of the p network. When constructed of precision resistors, a precise gain is 
retained over long periods of time. The price paid for this stability, of course, 
is a reduction in feedback system gain. 

Example: A range of ± 10 per cent is allowed for the internal gain of an 
amplifier, with A = 100. How can this gain change be reduced to ± 1 per 
cent? What is the resultant gain? 

We have AA/A = 0.10 and desire AA'/A' to be 0.01. Then 

1 



0.01 - 



X0.I0 



1 +Aft 
1 + A0 = 1Q 
Since A = 100, 

100^ _I0—I=9 
/? = 0.09 

We then have Ap = 9 for the feedback system. 
With 1 I AP = 10, the feedback system gain is 

.,_ A 100 _. , 

and we have reduced the gain by a factor of 10 in stabilizing the gain by a 
factor of 10. 

If we design our amplifier for a gain of 1000 ± 10 per cent, however, 
we have 

I 



and with A = 1000 



001 -TTAt 

I + AP = 10 



x 0.10 



1000;? = 9 

P = 0.009 






204 



Negative Feedback in Amplifiers 



and we have Afi = 9 again. But the feedback system gain is now 

1000 



A' 



1 +9 



- 100 



This is the desired gain but it is now stabilized to — 1 per cent. 

We have had to buy an amplifier with a gain of 1000 to obtain 1 per cent 
stability at a gain of 100. This, in certain applications, may be a small price 
to pay. 

Example: An amplifier is designed with A = 4000. Choosing /? = 0.04, 
we have 



Then 



A' = 



40 = 4000 x 0.04 = 160 
A 4000 



= 24.84 



1 + Afi ~~ 1 + 160 
Suppose the internal gain doubles to 8000. Then 

A0 = 8000 x 0.04 = 320 
8000 



A' = 



= 24.92 



1 + 320 
Gain A might drop to 2000, in which case 

Aft = 2000 X 0.04 = 80 

a ' = ttto- 24 - 69 

We have demonstrated that with A large, Eq. 9.1 1 holds and the gain can be 
maintained close to 

A'- 1 - ' -25 
regardless of large variations of the internal amplifier gain A. 



9.3 Bandwidth Improvement with 
Negative Feedback 

In Chapter 8 we showed that the high-frequency gain of an /?C-coupled 
amplifier was 



A "r(mld) 



(9.12) 



Bandwidth Improvement with Negative Feedback 



205 



Using a resistive fi network, the phase angle of /} can be held constant over 
the frequency band. For a feedback system, we could write 



A, 



[mldl 



,>_ a,™ ._, i + um) = 



^rlroli 



[i+Mo-d+g 



(9.13) 



We learned that we are at a limit frequency when the terms in the denominator 
are equal so 

-& - 1 + M«-id» 

and our upper limit frequency with feedback, designated f\, is 

n = [l + Mewl/. (9.M) 

The upper frequency limit of the system has been raised by (1 — Af}). 
Similarly, we can show 






'»-l+Mw« (9 ' I5) 

and the low-frequency limit is reduced by feedback. The bandwidth has been 
materially increased. 

Using f 2 as the bandwidth or assuming /' 2 > /',, the gain-bandwidth 
product with feedback is 

G-BW - , .-V 1 ' I' + fr*'«-«d/a = ^.w/ - . ( 9 -l 6 ) 

and the G-BW figure is seen to be independent of the feedback p. The curves 
in Fig. 9.3 illustrate this, being reduced in gain as the frequency band widens. 
It is apparent that we have traded gain for bandwidth. 

Physically large and expensive blocking capacitors can often be avoided 
when the mid-frequency bandwidth is expanded to lower frequencies, by use 
of negative feedback. 



Gain 
MB) 



A 



Figure 9.3 Effect of negative feedback on bandwidth. 



206 Negative Feedback in Amplifiers 

9.4 Reduction of Nonlinear Distortion 

Large-signal amplifiers, forced 10 operate in wide excursions across the 
device volt-ampere characteristics, will generate harmonics in the output 
signal. Negative feedback reduces such internally generated distortion of the 
the waveform. 

With the feedback loop open in Fig. 9.4, we have 

r a -AV',+ V k (9.17) 

where V h is the generator representing the internally generated distortion 
waveform. 



~© 



V! 




v: 



(9.19) 



Figure 9.4 Distortion inside the feedback loop. 

If we add negative feedback, the input signal is 

V\ r-V.-Vr (9.18) 

and the output signal is 

V, - AV\ + V h 
Substituting 

V = A<V,- V f )+V h 
and using Eq. 9.9 for V, with feedback, as well as V, = ftV'„ we have 

V' -A[(l ! Aft)V, - ftV B ] + V h 
Sorting the terms, we have 

v ;(i + Aft) = ,4(1 + ap)v; + v„ 

Making outputs equal, or V = V'„ we compare Eq. 9.17 without feedback 
and Eq. 9.20 with feedback. We see that the use of negative feedback has 
reduced the harmonic distortion by the factor 1/(1 + Aft), or 

D 



(9.20) 



Z>' = 



(9.21) 



i + Afi 

where D' and D represent per cent distortion. 

This result is of great importance in the design of high-power audio 
amplifiers. But with large Aft values the bandwidth is increased also and the 



Control of Amplifier Output and Input Resistances 



207 



high-frequency limit of audio amplifiers is frequently raised above 100 kHz. 
This occurs even though audio signals present no components over 20 kHz. 
Should Aft shift in phase angle in this extreme frequency range, there may be 
positive feedback and gain instability can be created. 

Example: An audio amplifier of 500 voltage gain produces 1 1 per cent 
harmonic distortion at full output. It was designed, considering C„ of the 
transistors, to yield an upper frequency limit of 8 kHz. 

Tolerable distortion is considered to be 1 per cent. What value of ft is 
needed to reduce the distortion and what is the bandwidth extension? 
From the distortion relation 

D 
~Tft 



D' = 



0.01 = 



1 + 
0.11 



1 + Aft 
\-r Aft=\\ 
and A ft = 10. With A = 500 we have 

The/j limit of the amplifier is extended by (1 - Aft) = 11 so that the/i 
limit with feedback is 1 1 x 8 kHz »■ 88 kHz. To improve the distortion 
situation, it is seen that we have extended the frequency range far beyond the 
needs of the audio signal. 

9.5 Control of Amplifier Output 
and Input Resistances 

For a constant input signal, the feeding back of a voltage sample V f , pro- 
portional to the output voltage of an amplifier, tends to maintain the output 
voltage at a constant value. This is done in Fig. 9.5(a) and constitutes voltage 
feedback. The amplifier appears to have a low output resistance. 

Similarly, the feedback of a voltage V f proportional to the load current 
tends to control the load current at a constant magnitude, independent of 
load resistance. This is the property of the circuit in Fig. 9.5(b) and represents 
current feedback. The amplifier appears to have a high output resistance. 

It is often desired to alter amplifier output resistances so as to supply a 
needed output current easily or to power match a load such as a loudspeaker. 
The two methods of obtaining feedback voltage provide the circuit designer 
with means for lowering or raising the output resistance of an amplifier. 

Considering voltage feedback first, as in Fig. 9.5(a), the output resistance 
R'o may be found by short-circuiting the independent signal source V,. Then 




208 



Amplifier 




Negative Feedback in Amplifiers 



Amplifier 




(a) (b) 

Figure 9.5 (a) Voltage feedback, scries input ; (b) current feedback, scries input. 

we apply a voltage V T at the output and R = V T jI T . With a short at V„ the 
input is V, = — RV T with the negative sign as a result of the subtraction of 
V f . From the currents in Jt„ 



and it follows that 



•H=4^'0«- 



R' = LZ — ■» 

° It II- A,.B 



(9.22) 



The output resistance of the amplifier is reduced by the introduction of 
negative voltage feedback. 

Referring to the current feedback circuit in Fig 9.5(b), we short V, and 
have V, = R f I T . Then 

V T . (A„R f i R | R f )l T 

With the feedback factor being 



we have 



Since R r < R 



B - Bj. 

P Rf + R* 



(9.23) 
(9.24) 



K '-G^ftc +I )^ +J ^ 



/f:=7 2: =/? (i i /*,/?) 

•T 



(9.25) 
(9.26) 



which shows that the output resistance increases by use of negative current 
feedback. 

The manner in which the feedback voltage is introduced into the input 
circuit can alter the input resistance of an amplifier. With the feedback volt- 



A Current Series-Feedback Circuit 



209 



age V f introduced in series, shown in Fig. 9.5, the input resistance with feed- 
back is 

R', = *,(1 + A,P) (9.27) 

and increases with B- 

When V f is introduced in shunt as in Fig. 9.6, the input resistance is re- 
duced as 

(9.28) 



R ' ~ 1 I A.P 



The circuit designer can choose /? and the method by which he derives the 
voltage V f to adjust the output resistance to a desired value. He can choose 
B and the method by which V f is inserted in the input circuit to alter the 
input resistance of the amplifier. These results are tabulated in Table 9.1. 











& 






















\ 


niplifier 












~- — n 


+ 


*'i 


|*<~> | 


+ 

K 























Figure 9.6 Shunt input of the feedback voltage. 



TABLE 9.1 Effect of Feedback on Amplifier Resistances 



Voltage derived : 
Current derived : 
Series input: 
Shunt input: 



R a decreases 
R increases 
Ri increases 
Ri decreases 



9.6 A Current Series-Feedback Circuit 

In Fig. 9.7(a) the unbypasscd emitter resistor R K provides a feedback pro- 
portional to load current; therefore it is current feedback. The voltage across 
R E is Vf and this is introduced in series with the input signal so that we have 
a current-series feedback circuit. That is, 

V„. = V,-V f (9.29) 

From the equivalent circuit 

V, ^ V bt + Vf = h,M •■ (/.-I •/,)** 

- [*,. + (1 + h f ,)R E ]h (9-30) 



210 



Negative Feedback in Amplifiers 





(a) 



<b) 



Figure 9.7 (a) Current feedback, series input; (b) equivalent circuit. 

Since V„ — — RI C = —h f ,Rl b , the gain can be obtained by use of Eq. 9.30 as 

K -h,,R 



A„< mii) y'-i^Q \h f .)R E 

By dividing out the h„ term and neglecting l < h f „ we have 



A,tnM\ = 



-g„R 



(9.31) 



(9.32) 



It is of interest to find that the feedback theory previously developed does 
apply to this practical circuit. 
The feedback factor is 



B = voltage feedback = (I + h, c ) R B / b ~ _Re 
p load voltage ~h f ,Rl b R 

Without feedback (R E = 0), the gain of the amplifier is 

A = -g m R 
We showed that with feedback the gain is 

r- A 
Using Eq. 9.33 and 9.34 we can form the feedback gain: 

-gmR -g m R 



(9.33) 
(9.34) 



A' = 



I+(-fc*>(-£) **■* 



(9.35) 



which is Eq. 9.32 written from the feedback equation. The negative sign on /? 



A Current Series-Feedback Circuit 



211 



appears because A v is negative for the one-stage C-E amplifier; this situation 
was previously discussed. 

Thus our feedback theory is confirmed. 

The input resistance can be obtained directly from Eq. 9.30 as VJI b , where 

« ( -5 I -*. + (i + h,.)R B 

= h„ + h f ,R E = A„(1 + g m R e ) (9.36) 

The series input of the feedback voltage has raised the input resistance from 
the /»„ value, present without feedback. Our theory says 

/?;-*,(' I Afi) (9.37) 

The input resistance is h u with no feedback (R B = in Eq. 9.36). Using the 
A and fi relations, 

K = /',.[l * <-*-*)(-if )] 

- A,.(l + g m R K ) (9.38) 

which is Eq. 9.36 derived from the circuit. 

To obtain the output resistance, we shall merely use Eq. 9.26: 

1 + A0 



R' = 



R 



since R is the output resistance at the 2, 2 port with no feedback. Using our 
A and /? relations, 

/?;-/?[l-f(-g m /?)(-^)] 

- *(1 + g m R B ) (9.39) 

with feedback. This is increased by reason of the current feedback. 

Example: For the circuit in Fig. 9.7, we use R E = 1.5 k£2, R = 10 kQ, 
h„ = 2 kSl, h f , — 50 and h ar — 10~* mho. Find the gain and input and out- 
put resistances, without and with feedback. 

We have 



gm 



_/'/.__ 50 



2(KH) 



= 0.025 mho 



Without feedback, 

A = -g n R - -25 X 10" 3 X 10 X 10' = -250 

R t = h„ = 2000 Q 

/?„ = /?= 10,000 Q (Mh ot neglected) 

With feedback. 



r_ -g m R _ 



-250 



1 +g m R B ~ 1 t (25 x 10-* x 1.5 x 10') 



= -6.5 



212 



Negative Feedback in Amplifiers 



To confirm the gain, we calculate 8 as 



'— ¥- 



1 500 



10,000 

and using the feedback gain expression 
A -250 



= —0.15 



A' = 



= -6.5 



1 + AB ~ 1 + (-250K-0.15) 
K = h,.(\ + g m R E ) = 2000(1 | 25 X 10" 3 x 1.5 x I0 J ) 
= 77,000 fi 

R' B = ' '.^ - 10 4 (1 + 25 X 10-' X 1.5 x 10 3 ) 
= 380,000 n 

5. 7 Voltage-Shunt Feedback Circuit 

The circuit in Fig. 9.8 gives shunt input for the voltage-derived feedback; 
therefore it is a voltage-shunt feedback circuit. 
A current summation at the base yields 

£ + // - h 



and we can write 



/, 



v. - r* 
R. 



K-t\.~y. 



R. 



l(MAAAr 




(9.40) 
(9.41) 
(9.42) 



lo- 



t-o2 



-02 lo- 



^O *i s 



£ 



-02 



Voltage Feedback with the FET 



213 



In writing Eq. 9.40 and 9.4I we have assumed that the internal gain A is large 
and V bf is therefore small. Likewise, in using /, = —VJR in Eq. 9.42, we 
have said that 7, < I c or that R, > R- 
Summing the currents, 

V. ■ V. _ Vq 
R, r R f h fc R 

V °\h fc R R f ) R, 

Then 

V„ I / I 

V=~R,[ 1,1 (9-44) 



(9.43) 



(a) (b) 

Figure 9.8 (a) Voltage feedback, shunl input; (b) equivalent circuit. 



and the gain is 



y* ' / ' \ 
y.~ *J_l + -L 



, _ V B R, h,.R 

A V, ~ R,R f -\h tc R 



Forhf.R » R„ the gain is determined solely by the feedback resistors R f 
and R„ as 

A'2Z-& (9.45) 

giving a very stable gain. This is equivalent to the result of Eq. 9. 1 1 and so 
we see that 

B- -£■ (9-46) 

for the shunt-feedback circuit. 

The result of Eq. 9.45 will be further discussed in Chapter 10. 

9.8 Voltage Feedback with the FET 

In the FET amplifier in Fig. 9.9, a portion of the output voltage is provided 
by the R„ /? 2 voltage divider and inserted in series with the input. A similar 
circuit is useful with the vacuum lube. This is a voltage-series feedback ap- 
plication. 

By opening R, the feedback is removed from the circuit and the gain is 

A, -g m R D (9.47) 

The sum R, + R z is made large with respect to R„ so that /, -C h- The 
current in the voltage divider is 

V. 



A = 



R, +R 2 



and the voltage feedback V f is 



R, + Ri 



(9.48) 



214 



Negative Feedback in Amplifiers 






The Emitter Follower as a Feedback Amplifier 

The gain with feedback is 
A' = 



215 



r4 f ' ' 



IO- 



+ 



Figure 9.9 An FET with voltage-scries feedback. 

This feedback signal is introduced as 

r,- K- V, (9.49) 

and the feedback is negative. The feedback factor can be obtained as 

(9.50) 



B =^-t = ~ R * 



This is an accurate result because there is no gate current to consider. 
The gain with negative feedback is then 



A' = 



TTAp 



-z»R D 



~gn,R D 



1 + *. 



Ri I Ri 



(9.51) 



Example: An FET amplifier has g„ = 0.004 mho, R D - 10,000 CI, 
Ri + Ri = 100 kQ, and /? = -0. 10. Find the gain with and without feedback. 
Without feedback, 

A, = -g m R D = -4 X 10' J X 10 4 = -40 
The feedback resistor R 2 is found from 

B = ^i 

P Ri+R* 

R, = -0(R l -I- R 2 ) = 0.10 X 10 s = 10* Q 



' ~g-*» - 40 

, H gm RoRi '7ZIZ in-J 10< x 1Q4 



'*. + R* 



1 + 4 x 10- 



W 



-40 

5 



Using Eq. 9.5 as a check, 



40 



A '' 1 H AB ~ 1 + (-40X-0.10) 



= -8 



3.9 The Emitter Follower as a Feedback 

Amplifier 

From the emitter follower circuit in Fig. 9.10, we have 

V bc =V,-V 

which is equivalent to 

V\ = V, - V, (9.52) 

for our general feedback amplifier. The result of Eq. 9.52 tells us that V t = 
V„ and the entire output voltage is being fed back to the input. This means 
that 

(9.53) 



>-£-&-' 



Since the output voltage is being fed back, the circuit gives voltage feedback 
with B = 1 and is extremely stable. 



Vcc 

? 



I o 1(- 



Vs R } 




|o- 



-\^>2 'v 



>'„f g„,K,(t) 



I c- 



'fr + M 



t'. 



-o2 



+ 



-o2 



(a) 



(b) 
Figure 9.10 The emitter follower. 




216 



Negative Feedback in Amplifiers 



Writing 



and 



V. = (h + I C )R E S h f< R E I b 

V, = h„l b + h f .R B l„ = (/.„ + A,,* e )/ 4 



we find the gain as 



4.~,-& 



A /f /?, 



gm* £ ~~ 



S I 



(9.54) 



' h lt i /; /r fl E l + gm R E 

which is the result obtained for the C-C circuit. The input resistance is in- 
creased by the series feedback and the output resistance is reduced to «= \/g n , 
as for the C-C circuit, which the emitter follower is. 

Similar results are obtained for the FET source follower and the triode 
cathode follower. 



9. 10 Multiple-Stage Feedback 

When we cascade C-E amplifiers, each stage adds a phase shift of 180°,' or 
n x 1 80" is the phase shift for /; stages. The polarity of the V, voltage must be 
maintained negative to V, for negative feedback and AR must be positive, 
cither as a result of ( i A)(+R) or (-A)(-R). 

With a single stage of C-E amplification, the output V is at 180" to the 
input signal and voltage feedback V, is introduced into the base, where it is 
negative to the signal as required. This is the method in Fig. 9.8. 

Because of the 360° phase shift of V with respect to V, in the two-stage 
C-E circuit of Fig. 9. 1 1 , the voltage feedback V, must be introduced into the 
emitter. This amounts to reversal of the V, voltage and gives ( \-A)(+fi) = 
AR. Thus we have methods for insertion of the feedback voltage with n odd 
or even. 




Figure 9.11 Voltage feedback, scries input, over two stages. 



Amplifier Cain Stability with Feedback 



217 



In combining C-E, C-B, or C-C stages or FET or tube equivalents inside 
the feedback loop, we must determine if the overall phase shift is an odd or 
even multiple of 180°. The proper entry point for the feedback voltage V t is 
then known. 

Feedback over multiple stages gives a larger value of A, resulting in a 
larger AR term; therefore, we can have greater reduction of distortion. There 
is some risk in amplifier stability because with increased numbers of capacitors 
inside the feedback loop, the phase angle of A may not be exactly 180n° at all 
frequencies in the response range. Conditions for positive feedback may be 
approached at the limits of frequency response, resulting in amplifier insta- 
bility. 

The major feedback in Fig. 9.1 1 is voltage derived but a small amount of 
current feedback is added in each stage by the unbypassed emitter resistors. 
Feedback of both types is additive in an amplifier and the effective R is the 
sum of the voltage and current feedback R values. 

Figure 9.12 shows a two-stage C-E current feedback amplifier. With a 
360° phase shift, the current supplied from R E opposes the signal current in 
R, so that the feedback is negative or subtractive to the input signal. 




Figure 9.12 Current-shunt negative feedback. 

9.11 Amplifier Gain Stability with 

Feedback 



We have used ideal feedback conditions as they occur in the mid-frequency 
range of an amplifier, requiring the feedback voltage V, to be opposite or at 
180° to the input signal voltage. We can design amplifiers to meet this condi- 
tion satisfactorily over a specified frequency range. 

We also encounter situations, however, in which the required phase angle 
is not present at extremely high or low frequencies, at which the angle of A, 
approaches 90/i° in RC amplifiers of n stages. At frequencies above f t , or 



218 



Negative Feedback in Amplifiers 



below/,, the voltage V t may not directly subtract from V„ and V\ increases. 
As a result the amplifier output is larger and the gain increases at frequency 
extremes, as shown in Fig. 9.13, and we have the condition of positive feed- 
back known as regeneration. 



('■am 

4» 



Mid-Range 




/ 

Figure 9.13 A feedback amplifier gain curve wilh regeneration at high and low 
frequencies. 

The dividing line between negative feedback and positive feedback ap- 
pears at 

\l + Afi\=l (9.55) 

and with 

\l+A0\<\ (9.56) 

the gain is unstable as a result of positive feedback. When the Afi value leads 
to 

\l + Afi\ = (9.57) 

or Afi — —I, the gain A' becomes infinite according to 

A A 





A' = 






1 + Afi 

and this is the extreme condition of gain instability known as oscillation. 
Using Afi as a stability criterion, we can write 

Stability: \Afi\>0 

Instability: > \Afi\ > -I 
Oscillation: Afi =- -1 

A Nyqujst plot may be used to study the action of Afi and is a polar 
plot of Afi at all frequencies from zero to infinity. Table 9.2 shows Afi 
magnitudes and phase angles for an RC amplifier, for A — 40, fi = 0. 1 in 
Eqs. 8.27, 8.28, 8.47, 8.48. Values from the table are plotted in Fig. 9.14. 
The frequency point moves clockwise around the plot; at 6 = +45° we have 
fi, at = —45° we have/ 2 . The mid-frequency range appears on the x axis 
at Afi — 4.0, since the phase angle is zero in the mid-range. Since Afi never 
becomes less than 0, the one-stage RC amplifier is unconditionally stable. 

Suppose that two such stages are used in cascade. At the frequency 
extremes the amplifier phase angles approach _L180°, and the Nyquist plot 



Amplifier Gain Stability with Feedback 



219 






y 


/ 3^ 


\/ 


i, i ii , 


-1 




^^\° 


\ ' 


IICj-KJ 






f 2 




X 



Figure 9.14 The Nyquisl plot. 



TABLE 9.2 A$ Plot of RC Amplifier 



m 


Ap 


8 (degrees) 


flh 


AP 


(degrees) 








+90 


Mid-range 


4.0 





0.1 


0.4 


84 


0.1 


= 4.0 


-6 


0.25 


1.0 


76 


0.25 


3.9 


-14 


0.5 


1.8 


63 


0.5 


3.6 


-27 


1.0 


2.8 


45 


1.0 


2.8 


-45 


2.0 


3.6 


27 


2.0 


1.8 


-63 


4.0 


3.9 


14 


4.0 


1.0 


-76 


Mid-range 


4.0 





00 





-90 



swings into the negative amplitude region. It may approach or surround the 
critical point at Afi — — 1. Such situations are shown in Fig. 9.15. 

The Nyquist criterion states an amplifier is unstable if the Nyquist curve 
encircles the — 1 point and is stable otherwise. 




Mid-Range 



Figure 9.15 (a) Nyquist plot for a regenerative amplifier; (b) an unstable 
amplifier. 



220 



Negative Feedback in Amplifiers 



By this rule we sec that the amplifier response diagrammed in Fig. 9.15(b) 
is unstable and will oscillate; that in Fig. 9.15(a) is regenerative due to the 
near approach to the — 1 point. For this amplifier the gain-frequency curve 
will have a high-frequency hump. 

In general, instability can be corrected by change of the Ap phase angle 
with a parallel R, C circuit in the line supplying V f to the input; or by reduc- 
tion of A through use of a series R, C circuit across a load, thus reducing gain 
at f 2 and above. 



9. 12 Gain and Phase Margin 

The Nyquist diagram requires much labor but simple frequency plots of Ap, 
in decibel magnitude and in phase angle, can be used for determination of the 
safety margin below the instability levels at A/} = dB and = 180°. 



to 







w 
O 
u 

1 

< 
8 



0" 


^N/2 


i 
i 

t 
i 

i 


1 
1 

* 

1 


-90° 




l 
l 
1 


1 
1 




Phase 

Margin 


* — 1 


1 

1 
v 1 


180° 


^^. 


">70° 


1 1 — i 1 i i i i 1 , , — .I.. ..j- 



10 



50 100 500 1000 

Frequency (kHz) 
Figure 9.16 Gain and phase margin. 



Review Questions 



221 



The gain margin is the decibel value of Afi at the frequency at which the 
phase angle reaches 180°. If negative, the amplifier is stable and can tolerate 
a theoretical gain increase equal to the margin, without regeneration. The 
phase margin is the angle of Ap at the frequency at which Ap reaches the 
0-dB level or the unity magnitude ratio. 

Figure 9.16 shows about —9 dB of gain margin and 45° of phase margin 
for a particular audio amplifier. Usual design limits are considered to be 
— 10 dB of gain margin and 30° of phase. The frequencies at which these 
margins arc measured are far beyond the normal mid-range of audio use; for 
instance, f t ^ 34 kHz but the gain margin is measured at 440 kHz. Again, 
this shows the necessity for phase control and gain limitation by decompen- 
sating circuits at frequencies well beyond the operating band in feedback 
amplifiers. It indicates that reduced distortion has been purchased at the 
expense of problems associated with stability in greater bandwidths. 



9.13 Comments 

Negative feedback is primarily used to reduce distortion in amplifiers and to 
make gain figures precise. We design excess gain into the amplifiers and sacri- 
fice this gain in using feedback in order to achieve desired results. With a 
large excess of internal gain, a large amount of negative feedback can be used 
and we obtain a precise overall gain without concern for variations of the 
internal gain figure. Widened bandwidth is an incidental result of using 
large amounts of feedback to reduce harmonic distortion. 

Summarizing the advantages obtained by the negative feedback process, 

1. Reduction of nonlinear distortion. 

2. Stabilized gain figures. 

3. Improved frequency response. 

4. Voltage feedback — lower output resistance. 

5. Current feedback — higher output resistance. 

6. Lower or higher input resistance. 

REVIEW QUEST/ONS 



9.1 Define negative feedback; regeneration. 

9.2 What is the feedback factor? 

9.3 What is meant by closed-loop gain ? 

9.4 Define voltage feedback; current feedback. 

9.5 What is meant by series-input feedback; shunt-input feedback? 

9.6 What conditions lead to increased gain with feedback? 

9.7 What conditions lead to decreased gain with feedback? 



222 Negative Feedback in Amplifiers 

9.8 In negative feedback, what is the phase relation of the voltage fed back to 
the input voltage? 

9.9 Equation 9.3 tells you whether the feedback is negative or positive. What do 
you look for? 

9.10 How does Eq. 9.52 tell us that the emitter follower has negative feedback? 

9.11 Under what conditions does feedback reduce distortion? 

9.12 What form of feedback increases the output resistance of an amplifier? 

9.13 What form of feedback reduces the output resistance of an amplifier? 

9.14 What is the circuit connection used to increase the input resistance of an ampli- 
fier? To decrease it? 

9.15 What is the effect of negative feedback on bandwidth? 

9.16 What is the definition of/?? 

9.17 When must /? be positive and when negative for negative feedback? 

9.18 How docs the connection of the feedback circuit of an amplifier differ with 
an odd number of C-E stages from that with an even number of stages? 

9.19 Why do audio amplifiers often have excessive bandwidth? 

9.20 How does feedback affect the stability of amplifier gain? 

9.21 How do you describe the form of feedback in an emitter follower? 

9.22 How is current feedback obtained in a triodc circuit? 

9.23 Why does an unbypassed emitter resistor reduce the gain of a C-E amplifier? 

9.24 Trace out the feedback loop in Fig. 9.1 1. 

9.25 Trace out the feedback loop in Fig. 9.12. 

9.26 What is a Nyquist diagram? 

9.27 Why should A$ avoid the value -1 ? 

9.28 Why is the mid-frequency range concentrated at one point in a Nyquist dia- 
gram? 

9.29 What is the Nyquist criterion for stability? 

9.30 Plot the Afi values of Table 9.1 ; show that a circle is obtained. 
931 What is gain margin? 

9.32 What is the phase margin? 

9.33 Name six advantages provided by negative feedback. 

9.34 How do we pay for the advantages of negative feedback? 

PROBLEMS 

9.1 For the black box feedback system in Fig. 9.17, if V, = 0.2 V, A = 20, and 
V = 1 V, find R, V,, V'„ and A'. 

9.2 For the circuit in Fig. 9.17, we have A = 50, 0.03, and V' = 5 V. Find 
V„ V\, and A'. 






Problems 



223 



9.3 For the circuit in Fig. 9.17, we supply V, - 5 V, with A = -20 and = 
— 1.0. Find the output V' , V,, V\, and A'. 

9.4 A l-V input signal for V, is used in Fig. 9.17 with A - 60, R = 0.07. What 
is V' , V f , and the gain A'l 



i; 



^d> 



v\ 



V. 



Figure 9.17 

9.5 An amplifier of three stages with voltage gains of —50, —15, and -10 has 
overall feedback applied with - -0.01. What is the overall gain with feed- 
back? 

9.6 What value of (3 is needed to reduce the gain of the amplifier of Problem 9.5 
to -100? 

9.7 In the circuit in Fig. 9.7(a), R = 10,000 £2, R E = 1000 £2, h„ = 1 700 £2, and 
g m = 0.007 mho. Find the voltage gain; find /?,'; find R'„. 

9.8 In the circuit in Fig. 9.8(a) we wish to incorporate 10 dB of negative feedback. 
Find R f if R, = 50,000 £2, R - 4000 £2, and A = -50. 

9.9 An amplifier has /l p(rold) = 200 and/ 2 - 50 kHz. When we add negative feed- 
back with ft =0.10, what is the mid-frequency gain and what value of high- 
frequency band limit do we obtain? 

9.10 In the circuit in Fig. 9.9, we have R„ = 10,000 £2, g m - 0.004 mho. With R, 
+ R t = 100 k£2, find R 2 to give a gain of -5. 

9.11 In the emitter follower of Fig. 9.10(a) we have R E = 5000 £2, g m = 0.0025 
mho, and h„ = 1200 £2. Find the value of V bl for V, = 1.0 V. What is the 
gain? 

9.12 An amplifier has a mid-frequency gain of 300 and/, - 500 kHz. We wish to 
raise the upper frequency limit to 5 MHz by the use of negative feedback. 
What gain will remain? What can you say about the gain-bandwidth product? 

9.13 An amplifier has A = - 100 and R, - 5000 £2. What value of should be 
used to increase the input resistance to 50,000 £2? How would you suggest 
that the feedback circuit be arranged? What is the gain with feedback? 

9.14 A transistor with h„ =- 1000 £2, h f , = 60 is used in a C-E circuit with R - 
2000 £2, R £ (unbypassed) = 700 £2. Find /? and the gain with the feedback 
present. 

9.15 Find the input impedance of the amplifier of Problem 9.14, with and without 
R E present. 

9.16 An amplifier without feedback has A, = -2700. With feedback, the gain A', 



Negative Feedback in Amplifiers 

is -97. What is the value of /? being used? How much feedback in decibels is 
used? 

9.17 Feedback of 15 dB is added to an amplifier with internal gain A 250. What 
is the value of fi required, and what is the gain with feedback? For the same 
output voltage, what will be the change in input voltage? 

9.18 An amplifier has A,. = -80 and V„ -= 100 V with 8 per cent harmonic distor- 
tion. We wish to reduce the distortion to 0.5 per cent. What value of fi should 
be used, and what input voltage is needed to give the same output as before? 

9.19 The gain of an amplifier without feedback is 100. The output resistance without 
feedback was 1500 fi. Plot a curve of output resistance with negative voltage 
feedback as a function of /? over the range /? = 0.005 to /? = 0.15. 

9.20 An amplifier has an internal gain of 37 dB and at 50-V output has 1 1 per cent 
distortion. Feedback is to be used to reduce the distortion to 1 per cent. 

(a) What gain will be obtained ? 

(b) What input signal must be supplied for the same output? 

9.21 An amplifier has an internal voltage gain A - 512 and an output of 12 V. 
Feedback is added until 0.95 V is required as input to give the same output! 
Find the fi being used. 

9.22 An amplifier in an electronic voltmeter must have a voltage gain of 100 within 
0.5 per cent but the internal gain may change as much as 12 per cent due lo 
component changes. Determine the value of /? needed to meet the guarantee 
and the needed internal gain. 

9.23 With negative feedback an amplifier gives an output of 10.5 V with an input 
of 1.12 V. When feedback is removed, it requires 0.1 5-V input for the same 
output. Find the value of /? and of the gain without feedback. 

9.24 An RC amplifier has three identical stages, with/, = 48 Hz, h = 140 kHz for 
each stage. The overall internal gain is -450 and fi = -0.05 is applied. Deter- 
mine/', and/j for the amplifier with feedback. 






10 

Integrated 
Amplifiers 



Circuits that eliminate the scries blocking capacitor arc said to be direct- 
coupled. The mid-frequency range is thereby extended down to zero fre- 
quency. More importantly, by removal of the bulky blocking capacitor the 
amplifier size is reduced and it becomes possible to integrate an amplifier 
circuit on a silicon wafer along with its transistors and diodes. Such mono- 
lithic construction leads to small size, high reliability, reduced cost, and 
offsetting of temperature effects. Problems arise because of bias voltage needs 
but these are met by use of dual power supplies or by special circuit design. 
By adding negative feedback the operational amplifier or so-called "op 
amp" has evolved. With this device we closely approach an ideal element 
having constant and controlled gain. In addition, high input resistance and 
low output resistance are achieved. Such gain elements can be cascaded 
without much concern for the effects of variable loads at the output. 



10.1 The Integrated A mplifier 

In Fig. I0.l we have the circuit of an integrated direct-coupled amplifier. 
With negative feedback added externally, the circuit becomes a stable 
general-purpose high-gain amplifier. 

The complete circuit contains 10 transistors, 2 diodes, and 16 resistors and 



225 



Sk 9°" 



or 




CH 




+ 



00 

♦— WW- 



t— WW 



-WW- 

aT 




—<ee, 



2 



_o- 



o 

♦-WW- 

<*: 

aT 
Mwv- 

j< 

o 



^ 



a 



a 




c» 





o. 



-ww 





a 

3 

o 









The Integrated Amplifier 



227 



is built on a chip of silicon approximately 1 mm 2 . This chip is enclosed in a 
12-terminal package about 1 cm in diameter and £ cm high. 

While the amplifier appears to be complex, the circuit functions can be 
readily explained. Transistors Q, and Q 2 constitute an emitter-coupled 
differential amplifier stage. Input to terminal 3 gives an output in phase with 
ihc input and A is positive; input to terminal 2 inverts the signal or gives a 
reversed phase output and negative overall gain, — A. Terminal 4 is at — V cc 
and serves as the ground connection. 

The outputs from Q, and Q 2 drive a second emitter-coupled differential 
pair at Q 3 and Q t . The output is taken from Q t and drives Q s , which supplies 
the base input for Q, . This transistor furnishes the amplifier output at 
terminal 9 as an emitter follower, used to ensure a low output resistance. 

Transistor Q it through its action on the currents of Q } and Q it cancels 
shifts in dc level that would occur with changes in dc supply voltage. A 
decrease in V cc causes a decrease in the voltage at the emitters of 3 and Q«. 
This negative-going change in voltage acts on transistor Q 5 with its emitter 
output going to transistors Q 1 and Q 9 , and less current passes these transis- 
tors. But less current in Q } , Q„ and Q t raises the collector voltages of these 
transistors, canceling most of the original downward shift from V cc . Tran- 
sistor Q 6 is a constant-current control for the first differential pair. 

The output circuit of Q 9 and Q l0 shifts the dc level at the output so that 
it is substantially equal to the zero level at 2 or 3 with zero signal. 

All transistors arc manufactured in the same operations and while Q 5 , Q s , 
g„ and Q 9 could be eliminated without reduction in the overall gain of the 
block, their presence adds little to the cost and much to the overall stability 
of output with varying supply voltages. Made as a monolithic element on 
silicon, the transistor characteristics are more nearly the same than when 
discrete units are assembled. The resistors are formed on the chip as well and 
the connection costs arc not increased by the circuit complexity. 

The two transistors of each differential pair will cancel common varia- 
tions. Since the two transistors are formed on the same chip with only a 
few thousandths of a millimeter separating them, temperature differentials 
between transistors are negligible and the effects of temperature changes on 
transistor characteristics are canceled. 

There is a great variety of such integrated circuits available. In general, 
they consist of four stages and Fig. 10.1 illustrates these. The first stage is the 
differential amplifier of Q, and Q 2 , with differential output; the second stage 
is a cascaded differential amplifier with sindc-ended output, composed of Q 3 
and Q t . Third comes an emitter follower Q b to lower the dc voltage level back 
toward that of the input, and the fourth stage is the output at £?, - 

A study of the elemental circuits employed and some of the applications 



226 



228 



Integrated Amplifiers 



of these "gain blocks" are the objectives of this chapter. Additional informa- 
tion on integrated circuit processing will be provided at the end of the 
chapter. 

10.2 The Differential Amplifier 

As can be concluded from the previous description, the differential amplifier 
is the basic element of the integrated amplifier. Drawn in Fig. 10.2, the 
differential amplifier circuit is symmetrical about the vertical dashed line 
a-a' and identical parameter changes on each side are balanced out. 

Direct-coupled amplifiers are unable to distinguish between changes in 
the dc value of a signal and temperature-caused changes in v BE , h FE , and the 
reverse saturation current and so the output current will drift with tempera- 
ture. In the differential amplifier the two transistors and the load resistors 
form a balanced resistance bridge at zero signal. A simultaneous change in 
v BE of the two transistors will change the collector currents; the voltages at 
A and B will change identically but the difference A — B will not be affected. 
Temperature drifts can be limited to effective input values of 3 /iW per °C 
(2 nV per °F). 

The circuit is well suited to integrated unit production because the 
simultaneously produced transistors and resistors will be matched closely. 
Equality of resistors is an important factor that can be easily achieved, 
whereas to produce resistors of an exact magnitude is difficult in processing. 

U & differential input voltage is applied, the inputs K, and V x will be equal 
in magnitude but of opposite polarity. With equal transistor parameters, one 
collector current will increase and the other will decrease. Voltages at A and 
B change up and down so that there is a voltage V„ between A and B. The 
changes in the respective collector currents are illustrated in the transfer curves 
in Fig. 10.3. Since the sum of the two currents remains constant, there is no 
signal voltage change across R E . 

If a signal is introduced to both transistors in a common mode, with both 
inputs being equally positive or in phase as an example, the collector currents 
increase identically and the bridge remains balanced with equal voltages at 
A and B. The output V„ remains zero. The current in R B does change with a 
common-mode input, however, and a common-mode signal appears across 
R E . 

The equivalent circuit is drawn in Fig. 10.2(b), in a form that emphasizes 
the bridge action of the amplifier. With identical transistor parameters we can 
write the circuit relations as 



F, = (R, + h u )l„, + R E I, 

Vi = (.R, + A,.)/*, + *,/. 

/. = /., +/„=£ h 



rJb, + h/Jb, 



(10.1) 
(10.2) 
(10.3) 




(a) 




W 



(b) 



Figure 10.2 (a) The differential amplifier; (b) equivalent circuit. 



229 



230 



Integrated Amplifiers 



c 



| 



12 
8 
4 


4 

-8 
- 12 





fcl 






'c 2 








\ 


/ 










\ 


/ 










/ 


\ 










/ 


\ 



















8 12 



4 4 

Inpul (mV) 
Figure 10.3 Transfer curve for a differential amplifier. 

Using Eq. 10.3 in Eq. 10.1 and 10.2, we arrange the results as 

V t - (/?, + hjlt, + h fl R E I bl + h f .R E l b , (10.4) 

V 1 = (R, + /,,.)/>, -1- h„R B I b , | h f ,R E I bl (10.5) 

Subtracting, we have 

Vr-V%= (*, + l'„)(h, ~ I J 
However, with balanced or matched transistors a change in I b , is a negative 
change in /„„ or / tl = — I h , and so 

V,-V z = 2{R, + AJ/ ti 



from which 



Similarly we can find that 



'•■ 2(R, + h„) 



I - v* - V * 

lb ' 2(R, - /;„) 



(10.6) 
(10.7) 



The voltage between A and fi at the collector connections is 

v. = hM* - hf.RJ,, 

Substitution of the current values from Eqs. 10.6 and 10.7 yields 



v.=- 



with a differential gain as 



A«- 



v x -v x 



I | f' 



_ 8«Rl 



1+A 

hi. 



(10.8) 



The Differential Amplifier 



231 



With y, and V 1 opposite in polarity, V x = — K 2 , we have an output V Q . If V x 
and K 2 are equal in magnitude and positive, the output is zero. 

We have stated that no signal voltage appears across fl £ with differential 
input voltage; confirming this is the absence of R e in the output voltage 
expression. 

It is evident from either Eq. I0.6 or Eq. I0.7 that the input resistance to 
either base is 



Hi = & =£ - 2(*. + h„) 



(10.9) 



This represents the series resistance of the path through the transistors and 
V x and V 1 ; the signal currents bypass the emitter resistor and it could be 
eliminated from the equivalent circuit because no signal voltage appears 
across it. 

When the output is taken from one collector to ground, the operation is 
said to be single-ended and the gain is one-half of the differential gain, as 



K„ (single-ended) — 



2 ' S 



(10.10) 



The differential amplifier circuit becomes an emitter-coupled phase inverter 
when the signal is applied to one input and the second base is grounded. The 
output voltages at 2 and 2' are of equal magnitude and 180° in phase. This is 
shown in Fig. 10.4. 



Vcc 
9+ 




-o2 




-o2 

+ 



-«2 



\R B 



V, 



EE 



Figure 10.4 An emitter-coupled phase inverter. 






232 



Integrated Amplifiers 



10.3 Rejection of Common-Mode Signals 

The differential amplifier tends to cancel effects common to its two sides and 
this action extends to signal voltages that are equally introduced in phase 
(not oppositely, as with differential plus and minus signals). Such common- 
mode signals are frequently caused by ac variations in the supply voltage or 
by voltages from stray magnetic fields in the ground or signal leads. The 
common-mode signal is introduced equally to transistors Q, and Q z as shown 
in Fig. 10.5 and these signals arc usually unwanted in the amplifier output. 

In Fig. I0.5 we have the two inputs to the amplifier as K rf for the dif- 
ferential input and V c for the common-mode input. With differential input 
the transistor voltages are equal but plus and minus, or V,, = — V lt . For the 
common-mode signal, V u = V lt and 

y c - v„ 
v c = v H 

By adding. 



V,= 



"„ + v„ 



(10.1I) 



With a pure differential signal and V,, = — V,„ the common signal is zero. 



Differential- (Z\y ( 
Mode Signal \~S d 




Common- 
Mode Signal 



Figure 10.5 Showing a common-mode signal. 



Rejection of Common-Mode Signals 



233 



As long as we have a perfectly symmetrical circuit, the common-mode 
signal will be canceled but we want to know how to maximize the rejection of 
common-mode signals when circuit unbalances occur. If a positive common- 
mode signal causes the transistor currents to rise, as would be expected with 
the npn units in Fig. 10.5, the voltage drop across R E also rises. This increased 
emitter voltage subtracts from the input signal and causes negative feedback, 
reducing the gain for the common signal. Differential-mode signals are not 
passed through R E and no differential voltage appears there. We usually want 
the common-mode signal to be rejected so that a high value of R E for feed- 
back is indicated. 

+ y cc 




Figure 10.6 Single-ended differential amplifier. 

We often wish to have one side of the output at ground potential and use 
a single-ended output circuit, as in Fig. 10.6. Because of the loss of circuit 
symmetry, a common-mode signal will produce some output, although there 
will be a substantial reduction in gain for the common-mode signal. A 
measure of the rejection of the common-mode signal in the output is given by 
the common-mode rejection ratio, defined as 

CMRR — differential-mode gain 
common-mode gain 



(10.12) 




The output consists of differential output and common output so that 

V.-AJTi + AjT, (10.13) 



234 



We can develop a useful relation as 

~ A « V <V -* cmrrF,) 



Integrated Amplifiers 



(10.14) 



Since A^Vj is the desired output, we readily determine the effect at the output 
by the additive term (l/CMRR)(K c /K d ). 

To obtain the CMRR for the circuit in Fig. 10.4, we split the circuit down 
the a-a' line; since R E appears in each half, it is shown as 2R E in Fig. 10.7. It 
becomes R E when paralleled in the actual circuit. Each half is a C-E transis- 
tor circuit with an emitter resistor 2R £ and the gain is 



A, c = 



—hf,R L 



h„ 4- R, — 2h fe R l 



-gm*L 



l+£+2g m R t: 



(10.15) 



With Eq. 10.8 for the differential gain we can calculate the CMRR value for 
the circuit as 

CMRR = 2g m « E (10.16) 

This result confirms the assertion that R E should be large for a large value of 
CMRR. 

Example I: With a transistor having g m — 0.0025 mho, what value must 
R E have to obtain a CMRR value of 40 dB? 
To obtain CMRR as a ratio, 

40= 20 log CMRR 

logCMRR = ^ = 2 

CMRR - 10 2 = 100 = 2g m R E 



from which 



*« = 2x2.5 ( "xl0-3 = 1 3 ! = 20 - 00QQ 



Example 2: We have input voltages of V,, = 50 piV and V,, = —50 /*V. 
The difference-mode gain of the amplifier is A„, — 1000 and the CMRR 
value is (a) 100 (40 dB); (b) 10,000 (80 dB). Calculate the output. 

We find 



Then 



V t - V u - V„ = 50 - (-50) - 100 jiV = 0.1 mV 

V - V « + V " - 50 ~ 50 - a 
2 2 U 



Rejection of Common -Mode Signals 



236 




+ o — vvft — H Vi Qi 

1 





-AMAi o + 



I 



Figure 10.7 Circuit analysis for CMRR. 



The differential signal output is 

V = A*V t = 10 3 x 0.1 x lO" 3 = 100 mV 
and we have zero common-mode output. 



Example 3: For the same amplifier we now have signals V u — 1 .00 mV 
I V„ = 
We find 



and V h = 0.90 mV. 



V t = V h - V,, = 1.00 - 0.90 = 0.10 mV 

as the differential signal. 

The common-mode signal is 

(a) With CMRR = 100, we use Eq. 10.14 to find 

1 V c 



.wxm(i 



100 



W) - ■» 



5mV 



For the same differential input as in Example I, the output has been increased 
9.5 per cent by the presence of the common-mode signal, 
(b) With CMRR = 10 4 , we again use Eq. 10.14 to find 

y„ = 10' X 0.10 (l -I -rgj^y) = 100.095 mV 

With the larger CMRR the error in the output signal caused by the common- 
mode voltage is only 0.095 mV, or about 0.1 per cent. 



236 Integrated Amplifiers 

10.4 A Constant-Current Circuit for R E 

For differential-mode signals we expect the current in R E to be constant. We 
would also like R E to be very large to give us a high value of CM RR. We can 
achieve these results by use of transistor Q y in the constant-current circuit in 
Fig. 10.8. 

Looking at the / 3 series circuit, 

(R 2 + /?,)/, + V D - V EE = 

after neglecting I B as small. If V EE > V D , then we have for / 3 






/,^ 



(10.17) 



Around the base-emitter circuit of Q 3 we can write 

R>L + Vbe, = *»/, + V„ 
and using I } from Eq. 10.17 

n i Rl' ee _i_ y y 

1 " _ R _ R ' ' D 'BE, 

If we select diode D to have a voltage characteristic equivalent to that of the 
base-emitter diode of g 3 , then V D = V BE , and 



(10.18) 



h = 



(10.19) 



RARi \ R 3 ) 

This expression for transistor current contains no transistor parameters and 
is a constant. 




Figure 10.8 A constant-current bias transistor. 






The DC Level Shifter 



237 



Transistor Q, is a current feedback amplifier and so has a large output 
resistance at the collector. This may approximate several megohms. This gives 
a very large effective R e value in Eq. 10.16 and a very large CMRR value. 

The value of V BE decreases about 2.5 mV per °C (1.5 mV per °F) but the 
diode can be chosen to vary similarly and the cancellation of V n and V BK in 
Eq. 10.18 can be achieved at all usual temperatures. As a result, /„ is inde- 
pendent of temperature. 

These circuits appear at Q 6 and Q 1 in the emitter leads of the differential 
amplifiers in Fig. 10. 1. 

A simpler but less accurate constant-current circuit is formed by a 
transistor with a fixed voltage reference applied to the base. The collector 
current is then I c = h rE I B and constant. The accuracy of current control is 
dependent on constant v BE and h n values. This is the function of Q 9 in the 
amplifier in Fig. 10.1. 

10.5 Voltage References 

We have just discussed an application in which a transistor was biased by the 
forward-voltage drop of a diode. The forward-biased silicon diode may be 
assumed as presenting a constant voltage, about 0.7 V, regardless of current. 
To obtain greater voltages, several diodes may be connected in scries. By 
choice of the temperature characteristic of the diode, temperature compensa- 
tion can be added to the function of a voltage reference diode, as shown in 
Sec. 10.4. 

These drops are smaller than can be obtained by use of a Zener diode and, 
more importantly, a diode such as D in Fig. 10.8 can be processed in the same 
steps and with the same materials as are required for the emitter-base junction 
of a transistor on the silicon chip. 

10.6 The DC Level Shifter 

With each stage deriving its base input voltage from the preceding collec- 
tor, the dc voltage level of each base rises with respect to ground, as one 
progresses through a direct-coupled amplifier. It is therefore necessary to 
shift the voltage level back down to obtain an output at which zero output 
voltage corresponds to zero input voltage. Using resistors and transistors, the 
<lc level shifter in Fig. 10.9 is well suited to integrated circuit production 
methods. 

Input is supplied to Q„ which operates as an emitter follower. Signal 
output is taken from Q„ also an emitter follower, to provide a low output 
resistance. 

Transistor Q 2 is a constant-current device, with its base supplied by a 
fixed reference source; therefore I c = h rE l B and is constant. With current l E 



238 



Integrated Amplifiers 
t+Yrr 



Input o—PT \j 



Voltage 
Reference 




Output 



Figure 10.9 DC level shifter. 

also constant, since / £ ~ l c , the drop in /?, is constant. Along the signal path 
shown we have 

y BE , + R\1b + y B E, = K constant volts 
We call AK,„ a change at the input and AK 0U1 the resultant change at the 
output. Voltages with no signal at input and output are K,„ and V oal . Then we 
can write through the signal path 

AK ln + K lB - K volts = AV oat + V OM 

A change in the input is transferred directly to the output as 

AK,„ - AK 0U , 

but the no-signal voltage level at the output is lower than that at the input 
because 



K„ - K = V m 



(10.20) 



A level shifter of this kind appears in Fig. 10.1, using Q„ and Q 9 with 0, o 
for output. 

10. 7 The Operational Amplifier 

When negative feedback is added to an integrated dc amplifier of large 
internal gain, we have an operational amplifier or op amp. It was originally 
given the operational name because of its use in performing the mathematical 



The Operational Amplifier 



operations of addition, integration, and differentiation but its applications are 
now much more general. The basic circuit is that in Fig. 10.10(a). 

The amplifier may have a number of stages and resistor R f , or some other 
circuit element, supplies voltage-shunt feedback around the gain element, 
stabilizes the gain, and lowers the output resistance. We define the internal 
gain as 

(10.21) 



A y . 



We shall require that gain A be very large as a condition of amplifier design, 
perhaps A > 10*. This permits us to say that V, = at the amplifier terminal 
at 1. To justify this statement, let A = 10 4 and V„ = 10 V; then V\ = 1 mV. 
With large A, we are also saying that A/} > 1 in the usual feedback amplifier 
expression. 

With V, = 0, however, the current to the amplifier must be negligible as 
well, or /; = at terminal 1. Then a current summation there gives 

/, » -h (10.22) 

With the voltage V\ as zero, these currents can be stated 

/ ^ 



Using Eq. 10.22 



and the gain with feedback is 






v. 



V, R x 

This is the basic gain relation of the inverting operational amplifier. 



(10.23) 



/, 



ww- 




A/WV- 



o+ 




o + 



o- 



<a) «>) 

Figure 10.10 (a) The operational amplifier; (b) the noninverling form. 






240 



Integrated Amplifiers 



The stability of the gain with feedback is dependent only on two resis- 
tances, or impedances in general, and the value of the gain can be easily 
adjusted. Note that the gain with feedback is independent of the internal gain 
provided that the internal gain is made large. 

Equation 10.23 is a result of A being large, as was Eq. 9.1 1 : 

I 



A'=* 



J 



(10.24) 



so we see that for the inverting operational amplifier 

Using an amplifier integrated on a silicon chip and adding two resistors 
and a power supply, we have a stable package of gain. The package will have 
high input resistance, low output resistance, and a wide and controllable 
bandwidth. The operational amplifier employs the circuits of the preceding 
sections in functions somewhat as shown by the typical arrangement in Fig. 
10.11. 

Because of the differential amplifier at the input, there arc two input 
connections available: one resulting in an inverted output or a gain —A' and 
the other giving a noninverted output and gain | A'. This adds further 
flexibility to the operational amplifier. When not used, the second input is 



Input 



Feedback 



Differential 

Amplifier 



Differential 

Amplifier 






Emitter 

Follower 



DC' Level 
Shifter 



Constant 
Current 



Output 
Amplifier 



■*- Output 



Constant 
Current 



Figure 10.11 Functions of an integrated operational amplifier, illustrating the 
dc levels. 



The Operational Amplifier 



241 



applied to the positive gain input terminal of the amplifier. The feedback 
is returned to the negative input terminal so that 



K,-t-^ 



to ground. There is a differential input V, - V { and the output of the 
amplifier is 

V ^A{V,-V f )^A(y,--^- T V^ (10.25) 



and the gain is 



A vr 



i + 



AR, 



(10.26) 



R, I R f 
If A ^> 1 as we have required, then the gain with feedback reduces to 



j, _ R, + R/ 

A —rT- 



(10.27) 



This approximates a positive equivalent to Eq. 10.23 and becomes equal 
when /?,>/?,. 

Example 1: The circuit in Fig. 10.10(a) is used for signal inversion, with 
A = -100,000, R, = 1000 a and R, = 10,000 Q. The gain is 



A' = -& = 



10,000 



= -10 



R, ~ 1000 

Since ft = — R,IR f = -0.1, we could use the accurate feedback expression 
-A -10 s -10 s 



A' = 



—a _ ~'» _ ~~' u _ _q goon 

1 + Aft ~ 1 + 10 5 X 0.1 ~ 1 |- 10* 



which shows that when A = — 10 s , we can certainly neglect I < Aft. This is 
equivalent to saying that V, = 0. 

Example 2: We use the circuit of Fig. 10.10(b) in the noninvcrting connec- 
tions. The gain is 

., _ /?, + R f _ 1000 + 10,000 _ , , n 

A ' ~~R, iooo uu 

Since R = i?,/(/?, t R f ) = 0.091 1, then the accurate feedback expression 
gives 



A'=* 



10 s 



= 10.999 



1 + Afi 1 + 9090 
which again confirms our requirement that A be large. 



242 



Integrated Amplifiers 



The Integration Operation 



243 



10.8 The Unity-Gain Isolator 

One of the simplest applications of the operational amplifier is the unity-gain 
circuit in Fig. 10. 12, useful in isolating one circuit from variations that may 
occur in a load circuit. 
From the circuit, 

r t +r t - v (io.28) 

But V\ S so that 

V, = V (10.29) 

and the output voltage follows the input signal, without a phase reversal. 
Since the output resistance is low, needed output currents can be obtained 
without loading an input circuit at 1,1, and the circuit is a buffer. 




10- 



02 

Figure 10.12 Unity-gain isolator circuit. 



10.9 The Summing Operation 

Several input voltages may be simultaneously operated upon in the opera- 
tional amplifier in Fig. 10.13. Current /, represents the sum of the three 
current components through R„, R b , and R c . That is, -/ 2 = /, and 



Rf Ro &b R c 



and so 



u 



r " + R/ b 



■%"•) 



(10.30) 



and the output voltage is equal to the weighted negative sum of the several 
inputs. 

If R„ = R„ = R c , then 

R, 



Vo = -%(V. 



V b + V c ) (10.31) 

If R f = R„, the output represents a negative summation of the input voltages. 




r 



Figure 10.13 A summing amplifier. 



10. 10 The Integration Operation 

By use of a capacitor in place of R f , the operational amplifier will perform an 
integration of the input voltage. An integrator circuit is shown in Fig. 10.14. 
The result of integration can be understood if we recall that the current- 
voltage relation for a capacitance, connected between V and ground poten- 
tial at 1, is 

K = lj'/ 2 A (10.32) 




(a) 



1 V 



u- 




1 



1 V 



(h) (c) m 

Figure 10.14 (a) Operational amplifier as an integrator; (b) input rectangular 
pulse; (c) output integral, \jRC = 1 ; (d) output integral, l/RC = 0.1. 



244 

Now we have previously shown that 

/■ = -h 

because /,' S 0. Since V\ = 0, we have said 

/ =£ 



Integrated Amplifiers 



(10.33) 



Because of Eq. 10.33, 



/ -- V < 



Substitution of this result in Eq. 10.32 gives 



(10.34) 



(10.35) 



as the output-input relation for the operational amplifier. The output voltage 
is the integral of the input, with a scale factor of — \/R,C. 
If we make /?, = 1 MSI and C = 1 fiF, we have 

R,C = 10 s X 10" 6 = Is 

and the scale factor is — 1 . Since RC has units of time, in seconds, we can use 
the RC scale factor to scale problems using a time variable. 

The integration operation is demonstrated on the step input in Fig. 
10. 14(b), the result being of ramp voltage form as the capacitor charges. If we 
had made R = 0.1 MQ, then l/RC = 10 and the result would plot toward 
— 10, reaching - I V in one-tenth of the time, so that time is scaled by 
changing RC. 

Figure 10.15 shows a multiple input integrator with different scaling 
factors, as might be used in an analog computer. 

The differentiating operation, opposite to that of integration, can be 
performed by using C for R, and a resistance as R f . We encounter noise 
problems in differentiation, however, and the operation is avoided. 




Figure 10.15 Use of different scale factors in integration of several signals. 



A Millivoltmeter 



246 



10.11 The Comparator 

One useful application of the differential amplifier is that of a comparator of 
the magnitudes of a signal voltage and a reference voltage. The voltage 
comparator in Fig. 10. 16 is basic in digital computer applications. Its action is 
demonstrated by the transfer curve of the differential amplifier in Fig. 10.3. 
When the input signal is slightly greater than the reference voltage, the output 
swings to saturation; when the input signal is slightly less than the reference 
voltage, the output swings to saturation on the other side and the output 
voltage reverses. 

A comparator is used in digital voltmeters, where the input voltage is 
compared with an internally generated ramp voltage. The cycles of an 
oscillator are counted from turn-on, with the count being stopped with a 
signal from the comparator at equality between the unknown and the ramp 
voltage. If the ramp wave rises at 10 mV per millisecond and the oscillator 
is at 10,000 Hz, the display will read 1 500 counts for an input voltage of 1 .5 V. 
Placement of the decimal point will cause the meter to read 1.500 V. 



Input o 




Figure 10.16 Basic comparator circuit. 



10.12 A Millivoltmeter 



In Fig. 10. 17 another application of an operational amplifier yields a 
niillivoltmeter. The instrument M may be one of 1-mA full-scale deflection. 
Since the operational amplifier gain expression yields 



R, '' 



(10.36) 



R>R 



V, 



So we have 

A- R f 
V, - R,R„ 

This is the basic equation for the circuit and states output current per volt of 



(10.37) 



246 



Integrated Amplifiers 



100 kfl 

AAIW- 




Figure 10.17 A millivollmcter using an operational amplifier. 

input. If we measure /„ in milliamperes, then V, must be measured in millivolts 
so we have units of milliamperes per millivolt. 

With the circuit values shown in Fig. 10.17, we have 

10' 



U 



1 
TO 



10 5 x 10 

so that we shall have a reading of I mA, or full scale on the instrument, for 
10-mV input to the amplifier. 

The full-scale value in millivolts can be readily changed to other millivolt 
values by changing /?, or R f . 

10. 13 Frequency Compensation 

As a feedback amplifier of many stages, the operational amplifier can have 
gain instability and oscillation at the high frequencies. The several stages have 
input capacitances for the transistors and the gain falls off or "rolls off" 
above limit frequencies fixed by these capacitances and associated resistances. 
Some amplifiers are internally compensated for stability but others require 
external R and C components so that the designer can choose his own 
bandwidth and stability margins. 

A typical uncompensated gain curve for an operational amplifier is 
shown in Fig. 10.I8, without feedback. The curve is drawn by use of the 
asymptotes of the frequency response curves, for simplicity in sketching. 
Uncompensated, the amplifier shows 60 dB of gain to the first limit frequency 
fi at about 200 kHz. Two more limit frequencies are determined by the RC 
combinations of the several stages at/' 2 and/:'. Each combination causes a 
rate of gain fall of -20 dB per frequency decade and these rates are additive, 
reaching -60 dB per decade above f% at about 20 MHz. 

We want to obtain 20 dB of closed-loop or feedback gain, thus using 40 
dB of feedback = (l - AR) dB. We draw the horizontal line at 20 dB and 
discover the gain to be limited at about 8 MHz and that the amplifier curve is 






The Slew Rate 



247 



Uncompensated 

/ 



20dB/Deeada 



40 dB/Uecade 



O 



-20 




- 60dB/Decade 



Frequency 
Figure 10.18 Uncompensated and compensated amplifier performance. 

falling at a rate of -40 dB per decade. From our study of feedback amplifiers 
we know that a rate of - 20 dB is associated with a phase shift of ±90° and 
is stable and that a rate of — 40 dB per decade represents a phase angle of 
_^180° with potential instability. From the discussion of gain and phase 
margin we know that our gain curve should cross the 0-dB gain line before 
the phase angle reaches 180°. For an amplifier to be stable, the gain curve 
should cross the 0-dB gain line at a slope not greater than — 20 dB per decade. 

We must compensate the gain curve by introducing a series RC circuit 
across appropriate terminals designated by the amplifier manufacturer. The 
roll-off of a simple RC circuit is — 20 dB per decade and we design the 
compensating circuit to have a corner frequency of about 7 kHz, sufficiently 
low that the -20-dB slope extends to dB before we reach the/', frequency. 
This is shown as the heavy line in the figure. The gain is less than dB before 
the phase angle of the gain reaches 180° at/', and the amplifier will be stable. 

Instead of reaching a bandwidth of 8 MHz, which might have been hoped, 
we now have a bandwidth of only about 400 kHz; however, this is still slightly 
wider than that of the amplifier alone. This is the price we must pay for 
stability. 

10.14 The Slew Rate 

Due to the charging time of the input capacitances of the transistors, the rate 
of rise of voltage is limited when a step input is applied to an operational 
amplifier. This maximum rate of voltage rise is called the slew rate. 

Consider the basic amplifier circuit in Fig. 10.10. The full input voltage 
step appears at input I because the output feedback from V cannot respond 






248 



Integrated Amp/Wets 



|f = Slew Rate 



Figure 10.19 Effect of the slew rate on rise or output voltage. 

instantly. This large value of input voltage drives one or more stages of the 
amplifier into saturation and the transistor capacitors are charged. As shown 
in Fig. 10.19, output V„ can rise only as fast as the internal capacitances can 
discharge and so we have the slew rate, given by 



5 = 



AK„ 



A/ 



(V///S) 



(10.38) 



This is a specification supplied by the manufacturer of the amplifier. 

The maximum rate of change in a sine wave is a function of frequency; and 
the amplifier slew rate may limit the amplifier in following a large signal at 
high frequency. The maximum rate of change of a sine wave occurs at the 
zero crossing and the slew rate should be greater, or 



^f > W. 



(10.39) 



where /is the frequency in megahertz, V mtx is the peak output voltage in 
volts, and the slew rate is in volts per microsecond. 

The slew rate is primarily of concern in selling a limit on the lime in 
which the amplifier may be switched from on to off. 

Example: Given an operational amplifier with a stated slew rate of 
10 V per //sand with a supply voltage of V^ = 15 V. In switching, the output 
will go from =c to = 15 V as on to off conditions. The time for this opera- 
tion is determined by the slew rale as 



'* S ~ 



15V 
10 V///s 



= 1.5^ 



With a sine wave of V a = 15 V peak value, the slew rate will limit the 
response of this amplifier to 



/- 



10 



TxV, 



o(max) 



2tz x 15 



-0.106 MHz 



If the peak of the sine wave is only 1 V, then the expression above shows that 
the response band would extend to 1.59 MHz. 






Definition of Terms 



249 



10. 15 Offset Voltage and Current 

With both input terminals grounded, a practical operational amplifier will 
develop a small output voltage, due to inherent imbalance in the circuits. The 
input offset voltage is defined as that voltage that must be supplied to one of 
the inputs to reduce the output voltage to zero. The value of the input offset 
voltage V„ can be determined from the output offset as 

^1 v 



r„ = 



(10.40) 



as in Fig. 10.20(a). The offset voltage V u is typically a few millivolts; hence 
amplifiers with appropriately small values should be selected. 

A difference in base bias currents at the input of an operational amplifier 
is called bias current offset. The effect can be reduced by insertion of a 
resistor R 2 in the circuit having excess current as in Fig. 10.20(b). The 
resistor should be 



R? 



R,+Rf 



(10.41) 



9 WW 




Q WW 




(a) (b) 

Figure 10.20 (a) Input offset voltage; (b) input offset current. 

10. 16 Definition of Terms 

There are a number of terms used in amplifier specifications that need to be 
understood, and we gather some of them here: 

Bandwidth. The frequency at which the voltage gain of the amplifier is 
3 dB below the voltage gain at mid-frequency. 

Common-mode Voltage Gain. The ratio of the ac voltage between the two 
output terminals to the ac voltage applied to the two input terminals con- 
nected in parallel for ac. 



250 



Integrated Amp/i/iers 



Differential-mode Voltage Gain. The ratio of the voltage change between 
two output terminals to the change in voltage between the two input ter- 
minals. 

Common-mode Rejection Ratio (CMRR). The ratio of the differential- 
mode voltage gain to the common-mode voltage gain. 

Differential Voltage Gain — Single-ended Output. The ratio of the change 
in output voltage with respect to ground at either output terminal to the 
change in voltage between the two input terminals. 

DC Dissipation. The total power consumed by the device under zero 
input signal and zero output conditions. 

Input Offset Current. The difference in the currents at the two input ter- 
minals for equal applied voltages. 

Input Offset Voltage. The dc voltage that must be applied between the 
input terminals to obtain equal quiescent voltages at the output terminals. 

Maximum Output Voltage. The maximum output voltage swing that can 
be achieved without distortion of the output waveform at the peak. 

Single-ended Output. An amplifier with output taken from only one of the 
two input terminals. 

Slew Rate. The slew rate is the maximum rate of change of output voltage 
with large signal inputs. 

10. 17 The Darlington Compound 

Transistor 

Two transistors can be assembled on a common chip in the manner of an 
integrated circuit, giving us a device called the Darlington compound transistor, 
shown in Fig. 10.21. 

By the usual transistor current relations, /, and / ; are 

But l bl is the emitter current of 0, so that 

h, - + h f ,)l, 
and so 

h = A/«(l + h f .)I, 



(10.42) 



(10.43) 






The Darlington Compound Transistor 



261 







Addition of Eq. 10.42 and 10.43 gives for the load current 
4 = [h fei + hfM + h fll )]l t 
The current gain is available from this expression as 



Figure 10.21 The Darlington compound transistor. 

(10.44) 



A, = -j- = h fet + hf„h f „ = h fe ,h fc 



(10.45) 



The last expression is simple and provides a fair approximation. 

The base-emitter circuit of Q 2 acts as the emitter resistance of g, and by 
use of the C-C relation we have 

R„ = K -I (1 i hfitom ( 10 - 46 ) 

Transistor Q z provides negative current feedback to Q i and raises the input 
resistance at 1,1. 

Using the relation between current gain and voltage gain as 

^loid 



A v = -A, 



*.„ 



put 



and Eq. 10.46, the voltage gain of the Darlington connection is 

R 



-h 



r " Hf "h lt , I ■hf.M.. 



hi,. 



(10.47) 



which is just that of a single transistor in the C-E circuit. 

By use of the Darlington connection internally made in the transistor, 
however, we have a single unit that gives the voltage gain of a single transis- 
tor, but with a materially increased input resistance. The Darlington com- 



252 



Integrated Amplifiers 



*V. 




Conslanl-Currcnt 
Transistor 



Figure 10.22 Darlington amplifiers added at op-amp inputs. 

pound transistor is frequently added at the differential inputs of an operational 
amplifier to raise the input resistance. This is indicated in Fig. 10.22. 



10. 18 The Cascode Amplifier 

As an example of the design of special circuits in integrated form, we have the 
cascode amplifier in Fig. 10.23. As simplified in Fig. 10.23(b), the signal 
circuit of Q x and Q } consists of a C-E transistor followed by a C-B stage. 
Transistor Q z is employed to vary the gain of Q 3 , through control voltage 
applied to terminal 10. 

The load of Q, is that of the emitter circuits of Q 3 and Q 2 . That is, 



'* 2(l+A Al ) = 2 ft 



(10.48) 



and is therefore small. The voltage gain -g m R,, of Q, is quite low. With a 
very low gain, however, the Miller effect cannot appreciably affect the 
capacitance at the input of Q x and so that transistor has a very large f z limit 
frequency and a large bandwidth. 

For the C-E stage the gain is approximately 

The gain of Q } , however, is that of a C-B amplifier 

A„ = g m ,R (10.49) 






Silicon Integrated Circuits 



263 




To Q 2 V cc 



|o- 



€ 




l(— o 



*h 



3 2 2 

(a) (b) 

Figure 10.23 (a) The MC 1550 integrated circuit; (b) simplified cascode circuit. 

The frequency range of a C-B amplifier is also wide, with/, = h fc f f , and so 
we have a circuit with three transistors on the same chip, in one package, with 
the input resistance of a C-E transistor but with a bandwidth of/, =/ r of 
the transistor. 

The complete package in Fig. 10.23(a) is used as a small-signal high- 
frequency amplifier. 

10. 19 Silicon Integrated Circuits 

The techniques of impurity diffusion from a gas, masking, oxidation, and 
metal deposition have made possible the production of complete circuits 
integrated into a chip of silicon. Reliability is high because interconnections 
are formed with the circuit elements. Size is drastically reduced and the 
response can be controlled because lead lengths and stray capacitances are 
constant. 

Figure 10.24 illustrates how the basic manufacturing techniques are 
applied in a simple integrated circuit. A />-silicon wafer, large enough for 
several hundred circuits, is the starting point. An n impurity is diffused 
through a photoresistant mask to form the n pockets in Fig. 10.24(b). The/wi 
diodes to the wafer will be back-biased and their resistances will isolate the 
circuit elements. Additional masking and etching provide holes in another 
coat of photoresist through which the p-base areas in Fig. 10.24(c) are 



254 



Integrated Amplifiers 
B C 



p Water 



(a) 







SiO, 




A>W p 

P 



H( )fei fesd! «1 



(d) 



C B E 



W 






Aluminium 

2 3 X 4 



SiO, 




(e) 



(0 
Figure 10.24 Steps in integrated circuit production. 



diffused. In the next step an n emitter is diffused into the base for a transistor 
at A. The result appears in Fig. 10.24(c). 

Finally the surface is oxidized, followed by etching of holes for the 
connections, and aluminum is deposited through a mask to form the leads as 
in Fig. 10.24(e). The electrical circuit is shown in Fig. 10.24(f). An npn 
transistor has been formed at A; a series resistor ofp material is at B, with a 
capacitance at C using the oxide layer as a dielectric. 

The steps in production are basically the same as those required to make a 
transistor and this is a production benefit. All transistors produced are 
virtually identical, although there may variations from one batch to the next. 
Diodes, resistors, and capacitors can also be formed and junctions given 
alternate usages as capacitances under back bias and resistances under 
forward bias, as well as serving as voltage references. 

When using discrete components, the active device, transistor, FET, or 
vacuum tube is considered a high cost item and resistors and capacitors arc 
used in preference. In the integrated circuit, cost is a function of circuit area 
and independent of the number of components per unit area and a transistor 
can be produced as cheaply as a passive element. New circuit designs become 
possible, using transistors to perform functions that might otherwise be 
carried out by passive circuit elements. The use of the constant-current 
transistor instead of a resistor in the emitter circuit of the operational 



passive Elements 255 

amplifier is an example of this. The constant-current circuit acts as an almost 
infinite resistance and greatly increases the common-mode rejection. An 
actual large R e would require an excessively large value of emitter supply 
voltage V EE and would waste power. 

10.20 Passive Elements 

A circuit capacitance can be developed between collector and base elements 
formed in the integrated circuit. There is also a capacitance in the reverse- 
biased diode to the wafer. The depletion layer in these junctions widens with 
increase in voltage; the capacitance decreases. Thus we have a voltage- 
variable capacitance. Capacitances as high as 1 000 pF per mm : are possible 
through the use of lightly doped collector materials. 

A thin film capacitor can employ the n emitter area as the under plate, the 
silicon dioxide coating as the dielectric, and aluminum or other metallization 
as the upper plate of the capacitor. There will also be a stray capacitance to 
the p layer, in the reverse-biased isolating junction. The silicon dioxide 
capacitance can develop 300 pF*per mm 2 . Higher values can be made by use 
of tantalum oxide as the insulating film. 

In Fig. 10.24 we used the bulk resistivity of one of the diffused areas as a 
resistor. A thin film resistor can also be vapor deposited on the silicon 
dioxide insulating layer, using nichrome or tin oxide. The surface is given 
another layer of silicon dioxide for resistance protection. 

The resistance of thin films is given by 



K A Wy 



(10.50) 



where L is the length, W is the width, and y is the depth of the resistance 
stripe. If L = W or, for a square, 



«-f 



(£2 per square). 



(10.51) 



The resistance R, is called the sheet resistance and has units of ohms per 
square. In general, where L and W are not equal, we have 



*-*. W 



(10.52) 



Using material having R, = 100 Q per square, we can make a 2000 Q 
resistor of 0.025-mm width as 

L 



2000 - 100 



0.025 



L = 



0.025 x 2000 
100 



= 0.5 mm 



Resistances in the range from 40 to 400 Q per square arc possible. 



266 Integrated Amplifiers 



10.21 Comments 

The operational amplifier is probably the most widely employed electronic 
circuit device. Available in several hundred integrated circuit forms, it has 
many hundreds of applications of which we have indicated a very few. 
While we have discussed some of the basic circuits from which an integrated 
package is assembled, we are not normally much concerned with, and cannot 
even sec, the tiny internal circuitry. It is the gain and phase characteristics 
between input and output terminals that are important to us; as long as the 
internal gain is very high, we are not greatly concerned with that figure. 

But with gain controlled and stabilized with negative feedback, with high 
input resistance so that it takes negligible current from the driving circuits, 
with almost zero output resistance so that it can supply any reasonable 
output current, the operational amplifier in an integrated package is very 
nearly an ideal electronic device. 

That the basic idea of the integration of devices on a silicon chip need not 
be restricted to complex circuits has been shown by the Darlington and the 
cascode circuits. Simple differential amplifiers are also available as single 
units or with as many as four units on a single chip. 

REVIEW QUESTIONS 

10.1 Why is the differential amplifier so well suited to integrated circuit production 
methods? 

10.2 What conditions of equality make the differential amplifier operate as a 
balanced bridge? 

10.3 What is meant by a differential signal? 

10.4 What is meant by a common-mode signal? 

10.5 What is ihe basic problem of amplification at dc or zero frequency? 

10.6 What is the purpose of a single-ended differential amplifier? 

10.7 Why is the differential amplifier circuit so well suited for dc amplification? 

10.8 What is the common-mode rejection ratio? 

10.9 Why is a large CMRR important in a differential amplifier? 

10.10 Can we have a circuit with single-ended input and differential output? 

10.11 What is the advantage of constant-current stabilization in a differential ampli- 
fier? 

10.12 What is the action of the constant-current circuit when there is differential 
input to the amplifier? 

10.13 What is the reason for the use of emitter follower circuits in integrated cir- 
cuits? 

10.14 What is an operational amplifier? 



■ 



Review Questions 



267 



10.15 Why do we use negative feedback in an operational amplifier? 

10.16 What is the ideal gain of an operational amplifier? 

10.17 What is an inverting amplifier? 

10.18 What is the gain in a noninverting amplifier? 

10.19 What is the ideal output resistance of an operational amplifier? 

10.20 Why do we use voltage reference circuits in an op amp? 

10.21 What is the function of a dc level shifter in an operational amplifier? 

10.22 Does V,al the input of an operational amplifier actually equal V? Why not? 

10.23 We sometimes say the input of a differential operational amplifier is at "vir- 
tual ground." What do we mean ? 

10.24 What is the value of the input current to an operational amplifier, in an ideal 
situation? 

10.25 What is the basic requirement in selecting an integrated amplifier for opera- 
tional amplifier use? 

10.26 Can you explain why the CMRR is infinite if a truly constant-current source 
is used in place of R E ? 

10.27 Sketch the transfer curve of a differential amplifier. 

10.28 Draw an operational amplifier in block diagram form. Explain each func- 
tional block that you include. % 

10.29 Define input offset voltage. 

10.30 At what rate of gain fall versus frequency do we expect to find instability in 
the operational amplifier? 

10.31 When we compensate an operational amplifier, what happens to the band- 
width? 

10.32 What components do we use to frequency compensate an operational ampli- 
fier? 

10.33 We wish to compensate an amplifier with an RC circuit having a corner fre- 
quency of 8500 Hz. We have a resistor of 25,000 £2; what C is needed? 

10.34 What is a unity-gain isolator? 

10.35 What is the purpose of an integrator? 

10.36 Draw a circuit that would subtract two voltages, using an operational ampli- 
fier. 

10.37 What is the slew rate? 

10.38 Why are the frequency and peak voltage related in the slew rate? 

10.39 In gain the Darlington connection is equivalent to what transistor circuit? 
What can you say about the relative input resistances? 

10.40 Why do we use a Darlington connection at the input of a differential ampli- 
fier? 

10.41 What are the advantages to be gained by use of a cascode circuit? 

10.42 With what circuit is the voltage gain of a cascode equivalent? 




Integrated Amplifiers 



Problems 



10.43 Why is the bandwidth of the first stage of a cascode circuit so large? 

10.44 What is the expected bandwidth of the second stage of a cascode? 

10.45 Name two methods of forming integrated circuit capacitors. 

10.46 Name two methods of forming integrated circuit resistances. 

10.47 What is the function of the SiO : layer in an integrated gain block ? 



PROBLEMS 

10.1 The circuit in Fig. 10.2 has R L = 10,000 fi, R E = 3000 fi, R, = 1000 fi, 
hi, = 1500 fi, and h { , = 60. Find the differential input-differential output 
gain. 

10.2 With h,, = 80, h„ - 1200 £2 in the circuit in Fig. 10.25(a), find the differ- 
ential voltage gain. 



- 15 V 



ww 1 





+ 15 V 



(a) 



(b) 



Figure 10.25 



10.3 A differential amplifier has a differential gain of 150. To find the common- 
mode gain we determine that V, = 2 V, V = 20 mV when the inputs are in 
parallel. What is the CMRR value in decibels? 

10.4 For a differential amplifier with R L 5000 fi, R E - 10,000 fi, R, = 2000 fi, 
h f , = 100, and g„ = 0.010 mho, find A«, A vc , and the CMRR value in 
decibels. 

10.5 Find V„ for a differential amplifier with single-ended output and V,, — 0.55 
mV, V,, = 0.40 mV, A« = 6000, and CMRR = 5 x 10*. 

10.6 Find the output voltage of a noninverting operational amplifier for V, = 
3.5 mV, R, m 40,000 fi, and R f - 240,000 CI, A = 50,000. 

10.7 Find the output voltage for the circuit in Fig. 10.25(b), with V, = -4 V to 
ground, if R t =0.1 Mfi and R, = 0.5 MCI, if A v - 10 5 . 

10.8 Find the output voltage for the circuit in Fig. 10.26(a), A = 10 4 . 



10.9 Find the circuit gain in Fig. 10.25(b), with A = 10 4 , R, - 100 CI, and R, = 
10,000 CI. 

10.10 Find the noninverting gain in Fig. 10.10(b), with A = 10 4 , R, = 100 CI, and 
R f - 10,000 CI. Compare your result to that of Problem 10.10. 

10.11 With R f = 500,000 Q, choose R, values and set up the circuit to obtain an 
output 

V„ = -iV„-6.SV b -AV C 

from three inputs V„ = 1 V, V b = 1.7 V, V, = 2 V. Hint: A unity-gain inver- 
ter might help. 

10.12 Find the output voltage for the circuit in Fig. 10.26(b). with A large. 



0.2 MSI 

- | Vo— WW — i 

0.5 MSI 
+ 2 Vo-^VWv 



0.5 Mfi 




- I VC- 



1 



T. r 




1 00.000 SI 



i 25.000 SI 



r 



(a) 



(b) 



Figure 10.26 



10.13 Find V„ if V, = 1 mV in Fig. 10.25(b), gain A = 50,000, R, - 100,000 £2, 
and R f = 50,000 SI. 

10.14 Find the output voltage for a three-input summing circuit with R f = 
0.2 Mfi, R„ - 0.2 MCI, R„ - 0.35 MQ, R c - 0.45 Mfi. V, = -2 V, V b = 
1 V, and V, = 3.2 V. 

10.15 Determine the gain A,,, - VJV, for the circuit in Fig. 10.27(a), using A, 





(a) 



(b) 



Figure 10.27 




260 Integrated Amplifiers 

= ^2 = 1 500. Remember that we can bisect the circuit into two circuits along 
the common line. 

10.16 An operational amplifier has internal voltage gain of 80 dB, and at high fre- 
quencies the gain rolls off at a rate of —20 dB per decade. Unity gain (0 dB) 
is reached at 2 MHz; feedback ft - 0.03. What is the gain A' in decibels, and 
what is the corner frequency? 

10.17 The/ 2 value of an operational amplifier occurs at 100 Hz and the no-feedback 
gain is 100 dB. What amount of feedback in decibels will be needed to reduce 
the gain to 40 dB, and what will the bandwidth become? 

10.18 Assuming the transistors are not identical in the Darlington compound circuit 
in Fig. 10.27(b), show that 

/?/, S A t R E 
A, ~ Iif„h/„ 
Find A,. 

10.19 Assuming the transistors of a Darlington pair, Fig. 10.27(b), are not identical, 
draw the circuit of an equivalent single transistor, and determine the h, and 
h f parameters in terms of h fe> , h ft „ h u „ and li lc ,. 



11 

Tuned and Video 
Amplifiers 



To select and amplify desired signals at radio frequencies we combine transis- 
tors or FETs with resonant circuits. Ideally wc would like to adjust those 
circuits to receive a narrow band of frequencies and reject all other frequencies 
but this is impossible at reasonable cost. So we define the selectivity of a radio 
receiver as a measure of its ability to separate a desired signal from other 
unwanted signals. In radio-frequency amplifiers we use a number of resonant 
circuits in cascade to raise the response to the desired band of signals and to 
reduce the response to unwanted signals near, but outside, the desired band. 
The transistor or FET isolates each resonant circuit and prevents the inter- 
action that would occur if these circuits were placed directly in parallel. The 
transistors also provide signal gain. 

Video amplifiers are designed with a frequency response extending from 
zero or near-zero frequency to 5 MHz or more for amplification of pulse 
signals. When designed with a response to 4.5 MHz, video amplifiers are used 
to amplify picture signals in television receivers. Transistors have gain fallofT 
in these frequency regions due to their inherent capacitance. We compensate 
the transistor circuits with inductance to extend the frequency range, again 
employing the principles of resonant circuits. 

11.1 Bandpass Amplifiers 

In the radio-frecpiency amplifier, the usual input signal consists of a center 
frequency /„ around which are grouped a band of frequencies. The center 

261 



262 



Tuned and Video Amplifiers 



frequency may range from one to many megahertz and the side frequencies 
may extend from a few to several hundred kilohertz on each side. 

Radio-frequency amplifiers are expected to provide selectivity for a desired 
frequency band and rejection of other frequencies. An ideal response 
curve is indicated by the dashed rectangle in Fig. 11.1(a). Ideally all fre- 
quencies in the desired band would be amplified equally and all frequencies 
outside the desired band, such as the interfering signal at/,, would produce 
zero response. 

We also require notability to shift the response rectangle over a wide 
range of frequencies at the will of the operator. Circuits must be simple 
since cascaded stages are simultaneously tuned and this requirement dictates 
that we depend primarily on the parallel-resonant LC circuit, with the capaci- 
tance element made variable for tuning. 

Such LC circuits provide only an approximation to the desired response 
rectangle and the discrimination against adjacent frequencies provided by the 
"skirt" portions of the curve is often inadequate. A measure of the relative 
selectivity is the shape factor of the response curve, as the ratio of the fre- 
quency width of the curve at 60 dB down from the center frequency response 
to the bandwidth at which a signal is only 6 dB down. 

For a single resonant circuit this shape factor may be 600 and, with 10- 
kHz bandwidth at 6 dB down from the resonant peak, the skirts cover 6 
MHz for the 60-dB suppression. Several parallel-resonant circuits can be 
cascaded with their associated amplifiers to reduce the bandwidth at 60 dB 
suppression. 

Other amplifiers operate at intermediate frequencies (I.F.) between the 
radio signal frequencies and the audio frequencies. For example, the usual 
I.F. in a broadcast receiver is 455 kHz. Such I.F. amplifiers are tuned to fixed 
frequencies and greater circuit complexity can be tolerated since the ad- 
justments are usually made only by the manufacturer. These complex cir- 
cuits provide wider bands and steeper skirts on the selectivity curve, as shown 
in Fig. 11.1(b). 




(a) (b) 

Figure 11.1 (a) Narrow-band R.F. response; (b) response of an I.F. amplifier. 



The Parallel -Resonant Circuit 



263 



11.2 The Parallel-Resonant Circuit 



The parallel-resonant circuit in Fig. 11.2 is the common frequency-selective 
circuit in tuned amplifiers. The resistance R p is usually that of the source, as 
shown in Fig. 1 1.2(b), where a transistor drives a resonant circuit. We con- 
sider the scries resistance of the inductor and the capacitor as small. 

At resonance the voltage V is in phase with the current / and the circuit 
at A, A' appears as a resistance; this is the definition of the resonance condition. 
To bring this about, the reactance of the capacitance must equal the reactance 
of the inductance: 

and 

ih =lnf ° L (,U) 

at the resonant frequency/,, derived from Eq. 11.1 as 

1 



/■„ = 



(2n) 2 LC 

1 



(« 1.2) 



InjLC 
A parameter Q is defined as the ratio of the parallel resistance to either 



reactance at resonance so 



R, 



<> ~ if = ^Jl - 2 « f ° CR > 

c 



= R,J^- 



(11.3) 
(11.4) 





(a) (b) 

Figure 11.2 (a) Parallel-resonant LC circuit; (b) with a transistor. 






264 



Tuned and Video Amplifiers 



A resonant circuit should be efficient in transmitting power to the next transis- 
tor or load. The power lost in the circuit is y : /R p since C and L are assumed 
to provide no loss. With R p inversely proportional to the power lost, then Q 
is also inverse to the circuit losses. That is, a large Q indicates a low-loss 
circuit and we consider Q as a resonant circuit figure of merit;'" Q values 
ranging from about 5 to 500 are found in such circuits. 
The impedance of the parallel circuit at A, A' is 



Z = 



R. 



v'+^Gb 






At resonance, X c = X L and wc have 

Z = R P 
Using Eq. 1 1.3 this can also be written 

Z„ - 2nf LQ = 



2nf C 



(11.5) 



(11.6) 



(11-7) 



Upon change of the frequency from the resonant frequency, either X c 
or X L increases and the difference term in the denominator of Eq. 11.5 be- 
comes larger and Z decreases. Since V Alt = IZ, the voltage falls as frequency 
departs from the peak value at R„, in Fig. 1 1.3. Driven by a transistor, the 
circuit provides a large voltage output at resonance and smaller voltages at 
all other frequencies. 

The phase angle of Z changes from inductive at frequencies below re- 
sonance to capacitive at frequencies above resonance. 

Equation 11.5 may be simplified for calculation purposes by rearrange- 




- O <t 



»«,= 









(a) (b) 

Figure 11.3 Equivalence of scries and parallel forms for resistance connections. 

'"When the circuit resistance is placed in scries with the inductance, the value of Q 
becomes 

The two equations for Q arc equivalent for the same circuit elements. 



Bandwidth of the Resonant Circuit 



ment. Consider the denominator difference term 



T c -Tr 2nfC ~WL 



-M"<**-=hd-m-f) 



(11.8) 



InfjLCl V L\f. f. 
after noting that 2n ^/LC — l/f . Inserting this result in Eq. 1 1.5, we have 

R. 



Z = 



We recognize that R P (C/L) = Q 2 from Eq. 11.4 and so have a simplified 
expression for the impedance 

Z = 



R, 



V' +«■(£-*)■ 



(11.9) 



In some cases, the resistance of the circuit is concentrated in series with 
the inductive branch, as in Fig. 1 1.3(a). For Q > 7, the series resistance R, 
can be transformed to a parallel R p , as 

», - r^ (11.10) 



R _ (2nf„LY 
~RT 



or 



R R7~ 



(11.11) 



If a parallel resistance already is present, then the effective R„ is the parallel 
value of the two. 

The various equations for the parallel RLC circuit, such as Q and band- 
width and resonant resistance, can be applied after R, is transformed to an 
equivalent R p . 

11.3 Bandwidth of the Resonant Circuit 

Figure II. 4 shows that the bandwidth of the resonant circuit, measured at 
the half-power points at which V = 0.707K„, depends on the value of R p . 
The exact nature of this dependence can be found by use of Eq. 1 1.9. Wc 
previously showed that the frequency of a bandwidth limit was found when 
the two terms of the denominator of an expression such as Eq. 1 1.9 were equal. 
Thus at the upper frequency limit/,, identified in Fig. 1 1.4, wc have 



.-<!■(£ -4) 



266 



Tuned and Video Amplifiers 



10 


















































































9 
























-r p = io.ooo n 
















































8 






























































V 




















7 


















'.W 


(l 


















































2 6 

o 


















































































> 

1 5 


























































/ 


/ 










-R 


p ■ 


5000 n_ 




& 4 


















7 






























/ 


Blf 




















3 


















































































2 


















































































1 






















































































































950 



975 



1000 
Kilohertz 



1025 



1050 



Figure 11.4 Resonance curves for two R p , L, C parallel circuits. 
Taking the square root. 

The lower limit at/, is below/, and the term in parentheses must be reversed 
to remain positive. After taking the square root, 

If/ 2 and/, are small departures from/,, then we can say that 



£i_f.~ 2(A-/o) 



(11.12) 



Suppose that/ 2 = I.I/,; then 



1.0 1.1 u.yi = u.z_ j 



Bandwidth of the Resonant Circuit 



and the approximation is reasonable. Therefore, we have 



■=M¥) 



» 

/, 



267 

(11.13) 
(11.14) 






Adding these expressions, we have 



/a-/, 






The bandwidth (BW) as/ 2 -/, is 

«P=/*-/,=^ (Hz) (11.15) 

This expression again defines Q and while instruments for measurement 
of Q are available, the value of circuit Q is often calculated from a mea- 
surement of the bandwidth of a circuit while it is mounted in the equipment. 
That is, /, and / 2 are measured where the voltage response is VJ^fT, and 
Eq. 1 1.15 is used to find Q. High Q and low losses are indicative of narrow 
bandwidth and large R„. By variation of the resistance R„ connected across 
the circuit, we can adjust Q and the bandwidth. 

Equation 1 1.4, 

shows that the C/L ratio of a circuit also controls Q, that is, a large ratio of 
C to L gives a high Q circuit. 

At the bandwidth limit frequencies the phase angle of Z is 45° below 
resonance and —45° above resonance at f % . 

Example: Choose L and C values for resonance at 1 MHz, with a reso- 
nant impedance of 100,000 Q. and bandwidth of 10 kHz. 
We then have Z„ = R„ = 100,000 £1 
Also 



Z„ = 2nfJjQ = R 



100,000 



6.28 x I0 6 x 100 
- 159 /zH 



= 0.000159 H 






268 



Tuned and Video Amplifiers 



We also have 

Q - 2nf.CR, 



C = 



Q 100 

2nf.lt, " 6.28 x 10 s x 10 s 
= 159 pF 



= 1.59 x 10-'° F 



11.4 The Single-Tuned Transistor 

Amplifier 

We frequently use a resonant circuit for both coupling and selectivity 
between two transistors, as in Fig. 1 1. 5. The load on the resonant circuit at 
2,2 is the input resistance of transistor Q 2 , usually increased over h lt by use 
of an unbypassed emitter resistor. 




Figure 11.5 Single-tuned, inductively coupled circuit. 

The secondary voltage V B , will be maximum when the circuit is tuned 
to resonance and 

I 



= 2nf.Lt 



(11.16) 



WJCt 

With M as the mutual inductance between the coils I, and L 2 , we choose 

2nf.M - ~/R^R7, (11.17) 

We find that the secondary voltage at resonance is 






(11.18) 



where our effective circuit Q, value relates to the circuit Q 2 of the secondary 
tuned circuit as 

Mi 



Q.= 



1 + 



(W.M) 1 



(11.19) 



The Double -Tuned Transformer 



Using the condition of Eq. 11.17, 



269 



and the bandwidth is 



0.-<b 



""-£-£ 



(11.20) 



which, with optimum M from Eq. 11.17, is twice the bandwidth of the secon- 
dary resonant circuit alone. By reducing M from the optimum value, we 
can control the bandwidth from twice the Q 2 value to that of Q 2 . 
The transformer should have the approximate relation 



= /Ru^n, 
V R n ~ n 2 



(11.21) 



where n,/n 2 is the ratio of primary to secondary turns. This is only an ap- 
proximate relation for such radio-frequency transformers, dependent on the 
completeness of the iron core surrounding the coils. 



11.5 The Double-Tuned Transformer 

For use in intermediate frequency amplifiers, where tuning is fixed by the 
manufacturer, we can widen and square up the response curve still further 
by tuning both primary and secondary windings of a transformer, as in Fig. 
11.6. Usually we make L, — L 2 and Q, = Q 2 = Q; both sides of the trans- 
former will also be independently resonant at/,. 

The magnetic coupling between L, and L 2 is measured by the coefficient 
of coupling k. With the coils far apart, k — and the secondary voltage is 
low. As the coils are moved closer together, the secondary voltage rises to a 
maximum at the value of critical coupling at k = k c in Fig. 1 1.6(b). As the 
coils are moved still closer, the secondary voltage falls again. The critical 
coupling value of A: is 

k c = -^== (11.22) 



and for Q t = Q 2 = Q we have 



V0.0* 



k. = -k 



(11.23) 



The circuits are said to be undercoupled if k < k c , critically coupled if k = k e , 
and overcoupled if k > k c . 

When we plot secondary voltage against frequency, and the value of 
k < k c , we have response curves with one peak, as shown for kjk c -= 0.5 
in Fig. 1 1 .6(c). The secondary voltage V., reaches a maximum peak value 
for k/k c = 1. This is equivalent to k = \/Q. Because of interactions between 



270 



Tuned and Video Amplifiers 




k/k c =kQ = 2.Q 




Figure 11.6 The doublc-tuncd circuit. 

the currents in the two resonant circuits when k/k c > l,k > l/Q, the res- 
ponse curve splits and shows two frequencies of maximum response and the 
bandwidth widens. 

The frequencies of peak response are 

I 






(11.24) 
(11.25) 



At critical coupling, with A: = l/Q, the expressions indicate that /„ =/» — 
/„ for the single peak, as they should. 

If the dip in the center is allowed to fall to 0.707K pe , k , the bandwidth is 

The gain at either frequency of peak response is 



M.. t | = 27t/ °^ gL; 



(11.26) 



with L, = L t . 

Such transformers cannot simply be aligned by tuning for a peak secon- 
dary voltage with overcoupling. One procedure is to reduce the Q value, 
thereby changing the coupling from overcoupled to undcrcoupled. This is 



The Pulse Waveform 



271 



done by shunting fixed resistors on both primary and secondary. If k — 0.05 
and Q, — Qi = Q = 50, then k t — 0.02 and the transformer is overcoupled. 
If we reduce Q to 15, then k c = 0.067 and the transformer becomes under- 
coupled and gives only one response peak. Tuning for this peak will correctly 
establish both primary and secondary resonance. The loading resistors can 
then be removed. 

The use of two tuned circuits produces a response over a greater band- 
width and with steeper skirts on the response curve so that the response 
more nearly approaches the ideal rectangle. More elaborate circuits are 
available with greater numbers of tuned circuits. Each circuit adds a small 
ripple of gain to the passband response but steepens the falloff at skirt fre- 
quencies. These are known as stagger-tuned amplifiers. The analysis is beyond 
the scope of this text. 



11.6 The Pulse Waveform 

In digital computation, data transmission, and radar we use the pulse wave- 
form, shown in ideal form in Fig. 1 1.7(a). After transmission through various 
circuits and amplifiers, the waveform may appear more as in Fig. 1 1.7(b) or 
(c). 

The pulse starts at / — but the received pulse is delayed by the transmis- 
sion path as in Fig. 1 1.7(b). The exact starting time involves guessing as to 
when the received wave leaves the zero axis; to avoid this uncertainty the 
rise time of the pulse is arbitrarily defined as /„ from 10 to 90 per cent ampli- 
tude. The pulse width is also difficult to determine so it is defined as t w , mea- 
sured between the 50 per cent rising level and the 50 per cent falling level. 
The top of the pulse may sag some percentage d or may overshoot by some 
percentage of the amplitude as shown in Fig. 1 1.7(c). 

One of the major needs for the video amplifier arises in the amplification 
of pulses because of the large bandwidth needed for good waveform repro- 
duction. A general pulse train, as used in data transmission, is shown in Fig. 




(a) 



(b) (O 

Figure 11.7 Pulse waveforms. 



272 



Tuned and Video Amplifiers 



m 



Figure 11.8 A pulse train. 

1 1.8. Each pulse is At seconds long and the pulses are repeated at a repetition 
rate off R = \/T R per second. Each pulse is made up of an infinite number of 
harmonic frequencies, starting with the fundamental or first harmonic at 
f R , a second harmonic at 2f R , and more harmonics at nf R frequencies to in- 
finity. Of course, the very high-order harmonics are small in amplitude and 
are cut off by the amplifiers since no amplifier can have a passband to infinite 
frequency. 

But the fidelity of the pulse waveform is dependent on the video amplifier's 
providing a uniform gain passband for these harmonics. Excellent reprpduc- 
tion is possible if the passband is wide enough to amplify frequencies to 




^ 



rfs 



fh) 




Figure 11.9 Approximaiion lo a square wave: (a) fundamental frequency; (b) 
fundamental plus third harmonic; (c) fundamental plus third and fifth harmonics. 



The Shunt-Peaked Video Amplifier 



273 



4/Af, and reasonable reproduction is obtained with a passband reaching to 
|/A/ hertz. Thus for a l-/is pulse equal to At, repeated 400 times per second, 
the bandwidth for excellent reproduction would be 4 MHz and for reasonable 
reproduction we would need a 1-MHz bandwidth. In that 4-MHz passband 
there are harmonic signals every 400 Hz, or 10,000 signals must be amplified 
equally. 

The process of building a wave through addition of the harmonic fre- 
quencies is shown in Fig. 1 1 .9. Very high-order harmonics arc needed to build 
the square corners. The harmonic amplitudes that are needed for the first 
few frequencies of a square wave are given in Table 11.1. 



TABLE 11.1 Relative Amplitudes of Harmonics in a Square Wave* 

Harmonic Amplitude Harmonic Amplitude Harmonic Amplitude 



1 


0.636 


II 


-0.058 


45 


0.014 


3 


-0.212 


13 


0.049 


55 


-0.011 


5 


0.127 


15 


-0.042 


65 


0.0098 


7 


0.091 


25 


0.025 


75 


-0.0085 


9 


0.070 


35 


-0.018 


85 


0.0075 



•Wave amplitude -- 1. Negative signs indicate reversed phase. 



11.7 The Shunt-Peaked Video Amplifier 

The video amplifier has been defined as having a frequency range extending 
from near zero to a few megahertz. This takes the operating range into the 
frequencies at which the Miller-effect capacitance of the transistor or FET 
normally produces a fall in gain. 

One way to widen the frequency response band of an amplifier is to reduce 
the load resistance. This reduces the gain and the Miller-effect capacitance 
and raises the/. limit of the amplifier. The/ 2 limit may be further raised by 
use of an inductance in the load. The resulting resonant effect raises the load 
impedance at those frequencies where C,,, in shunt, reduces the load. The 
gain is maintained uniform to a higher/ 2 value. 

In Fig. I l.lO we show the compensating inductance L, usually a fraction 
of a millihenry, connected in series with the FET load resistor R,. In Fig. 
1 1. 10(b) we have the equivalent circuit of the transistor, including the Miller- 
effect capacitance of Q 2 and the wiring capacitance, which together form 
C T . We have neglected and dropped the series blocking capacitor and the 
bias network from the circuit. 

With L = we have an uncompensated amplifier with load R, and the 






274 



Tuned and Video Amplifiers 



<?+ V. 



lo- 




«fe 



lo- 



ftl _£. 






<■< 



:« 2 K,, 



" 6+l',r- 




*+«'«■ 



(a) 




r- ^ 



a: - 



(b) 



Figure 11.10 (a) Shunt compensation by L; (b) equivalent circuit. 



upper limit frequency is 



h = 



I 



SCT <"- 27 > 

obtained by the methods of Chapter 8. We use this frequency as a norm, 
against which the higher/', of the compensated amplifier will be compared. 
The circuit is said to be shunt-compensated by the inductance. 
We define a parameter as 



_ 2nhL L 
H R, R;C T 



(11.28) 



after using Eq. 1 1.27. This equation expresses the ratio of inductive reactance 
at the/ 2 frequency to the series resistance. 

The inductance L is then varied and frequency-response curves plotted 
as in Fig. 11.11. The curve fore/ m is for the uncompensated amplifier with 
L = 0. The condition of q = 0.41 is found to give the widest frequency band 
without a rise of gain above the mid-range value and for that value of q the 
bandwidth is increased to/i = 1.72/. 



The Shunt-Peaked Video Amplifier 
+ 5 



275 



33 



-5 

















. io 




























a 

0, 


1 "^ 
).s\ 


























ti^^^X 






:odB 
it Decat 
























- p 


k- 












1 


= 


y 


































I r 













0.1 0.2 







0.S I 2 5 

Mi 
Figure 11.11 Shunt-peaked circuit response versus parameter </. 

Values of q > 0.41 widen the response band further but introduce a peak 
at some frequency less than/ 2 . It should be noted that while one stage may 
show a peak of G decibel, n stages will increase that peak to nG decibels and 
such a high peak may not be acceptable. 

We can find the time delay in transmitting a frequency through the am- 
plifier from the phase angle 6. Using 0/360 as the fraction of a cycle by which 
the signal is delayed and l//as the time of a cycle, we can write 

9 



time delay — 



360/ 



Figure 1 1 . 1 2 is plotted in terms of time delay. If all frequencies are given an 
equal time delay, the curve will be flat and the whole wave will be delayed 



0.16 




0.1 0.2 



0.5 1.0 2 5 

Mi 
Figure 11.12 Time delay for the shunt-peaked amplifier. 



276 



Tuned and Video Amplifiers 



The Series-Peeked Video Amplifier 



277 



but is unchanged in waveform. Therefore, q = 0.32 offers the best phase 
response. A compromise between q ^ 0.41 and q = 0.32 is often made at 
about q = 0.35. 

With </ = 0.35 and R, chosen for suitable low-frequency gain and band- 
width/,, we can find the value for the compensating inductance from Eq 
11.28: 

L = qR)C T (11.29) 

The high-frequency gain of the compensated amplifier is given by a com- 
plicated expression: 

/U " = - g "4 i-Hi-w(^ g W 2 )J 2 (,, - 30) 

where — g m R, is the mid-range gain. The phase angle expression is also com- 
plicated: 

0- -tair'£[l -q+q^L^ (11.31) 

The rise in gain at the high frequencies is due to resonance between L 
and C T , the lower peaks being produced as a result of a low value of circuit 



Example: An FET has g„ = 0.003 mho and is to be used in an amplifier 
to provide a mid-frequency gain of 10, with the gain extended to the highest 
possible frequency. The FET has C, , = 1 pF and C, d = 2 pF. 

We find R t from the mid-frequency gain: 

^.(midi -' —g m R, = 10 
10 



K.= 



(1.003 



= 3300Q 



The Miller-effect capacitance is 

C„ = C„ + (1 + g m R,)C, d 

= 1 + (1 - 0.003 X 3.3 X 10 3 ) X 2 
= 1 + 22 = 23 pF 

We then add 2 pF for wiring capacitance so that C T — 25 pF. 
The limit frequency of the uncompensated amplifier is 

1 1 



h- 



2nC T R, In X 25 x 10"' 2 x 3.3 x 10 J 
= 1.93 MHz 



By use of 



L = qR)C T = 0.41(3.3 x 10 3 ) 2 x 25 x 10" l2 
= 1.12 x 10-«H-=0.112mH 



we can extend the frequency range to 

f\ = l.72/ 2 = 1.72 X 1.93 = 3.32 MHz 

11.8 The Series-Peaked Video Amplifier 

A more elaborate circuit that also raises the high-frequency response limit is 

the series-peaked circuit in Fig. 11.13. We add a small inductance in series 

with C r , with which it resonates to raise the gain at the high frequencies. 

For the uncompensated amplifier with L -— 0, the mid-frequency gain is 

A c , mU , - -g m R, (H-32) 

and the/ ; frequency limit is 

1 



A- 



(11.33) 



2s(C, i C T )R, 

where C, is the output and wiring capacitances of the first transistor Q t . The 
optimally flat response is obtained when 



and 



q = MlL =, 0.67 
R, 



c, -^ 



giving a bandwidth of/i =- 2/ 2 . 
Using Eq. 1 1.33 in 1 1.34, we have 



L = A = qRKC, ■■ C T ) 
InJz 



(11.34) 



(11.35) 



(11.36) 



Often the output capacitance of Q, is not exactly one-third of C T , as called 
for in Eq. 1 1.35. To increase either C, or C T to obtain this ratio will reduce 
/, as calculated by Eq. 1 1.33. But a reduction in f 2 is not the result desired. 
Accordingly, the shunt-peaked circuit, with its lessened complexity, is gen- 
erally used in video amplifiers. 

-It- 



-CI 




+ v, 



cc 



Figure 11.13 The serics-pcaked circuit. 









278 



Tuned and Video Amplifiers 



'ova 

*u (mid) 



1.2 

1.0 
0.8 
0.6 
0.4 
0.2 



























C^ 


k 








K 


^ 










^ 
















0.1 



0.2 



0.5 



10 



Ml 



Figure 11.14 Comparison of shunt and scries compensation. 

A comparison of the uncompensated and compensated responses is 
made in Fig. 11.14. 



11.9 Review 

Q is a most important and useful parameter of the parallel-resonant circuit. 
It is defined as 



where 






i 



InJLC 

for the circuit-resonant frequency. 

We then know that the bandwidth is 

«r -/»-/.-<§ 

and that at resonance 

Z « R, = 2nf LQ 
For frequencies near resonance, 

R, 



Z = 



V'^tt-i)" 



For the tuned coupled circuit the bandwidth can be increased to 2/JQ 
with critical coupling k c for the coils and cuM equal to As /R i ,/?,,. Further 



Review Questions 



279 



widening and steepening of the skirts is possible by tuning of both primary 
and secondary and making k > k c . These methods lead to a response that is 
closer to the ideal rectangular shape for the response curve. 

The shunt-compensated circuit is most usually employed in video am- 
plifiers. 

REVIEW QUESTIONS 

11.1 Define resonance. 

11.2 Why do we desire a rectangular frequency-response curve? 

11.3 What is the condition of the circuit reactances at resonance? 

11.4 What is the resonant frequency? 

11.5 Define selectivity. 

11.6 What is the shape factor of a resonant circuit? 

11.7 What range of bandwidths do we expect to find in radio-frequency amplifiers? 

11.8 How is Q defined in terms of circuit elements? 

11.9 How is Q defined in terms of bandwidth? 

11.10 How do we often measure Q1 

11.11 What is the resonant resistance of a parallel circuit? 

11.12 Why is the resonant resistance a maximum at resonance? What happens to 
the currents in L and C? 

11.13 What is the bandwidth of a resonant circuit? 

11.14 A circuit has a resonant frequency of 100 MHz and a bandwidth of 5 MHz. 
What is the Q1 

11.15 A circuit has a shape factor of 100: I and is 6 kHz wide at 6 dB down from 
the peak. What is the width of the curve at 60 dB down? 

11.16 What are the phase angles of the resonant impedance at the bandwidth limit 
frequencies? 

11.17 How does the C/L ratio affect the bandwidth? 

11.18 What is the tuning result of 

(a) Critical coupling? 

(b) Overcoupling? 

(c) Insufficient coupling? 

11.19 In a tuned coupled circuit, on what factors does V„, depend? 

11.20 On what factors docs the bandwidth depend in a tuned coupled circuit? 

11.21 What is the maximum value of the coupled Q, in terms of the circuit Ql 

11.22 Define critical coupling in terms of Q. 

11.23 What is a video amplifier? 

11.24 How do we shunt compensate a video amplifier? 



280 Tuned and Video Amplifiers 

11.25 How do we scries compensate a video amplifier? 

11.26 Why do we avoid a gain peak in a cascaded shunl-compensated amplifier? 

11.27 What is the parameter q in a shunl-compcnsated amplifier? 

11.28 What is a good compromise value for q in a shunt-peaked amplifier? 

11.29 What is the requirement on the two capacitances in a series-peaked amplifier? 

11.30 How would you tune a double tuned overcoupled transformer? 

11.31 In a parallel-resonant circuit, in order to make the circuit inductive, should 
the generator frequency be raised or lowered from resonance ? 

11.32 To raise the resonant frequency of a parallel circuit, which is it necessary to 
do? 

(a) Increase the capacitance 

(b) Increase the resistance 

(c) Decrease the inductance 

(d) Increase the inductance 

11.33 What is the repetition rate? 

11.34 How is pulse rise time measured? 
1135 How is received pulse width measured? 

11.36 What is overshoot of a pulse? 

11.37 How many harmonics are in a pulse wave? 

11.38 How many harmonic frequencies are needed for excellent reproduction of 
a pulse wave with repetition rate of 1000 per second, at 10-MHz frequency? 



PROBLEMS 

11.1 A circuit is resonant at 2000 Hz. If the coil is 0.120 H, what capacitance is 
being used ? 

11.2 A parallel circuit is resonant at 5.35 MHz and uses a 40-pF capacitor. What is 
the value of the inductance? 

11.3 In a parallel-resonant circuit the reactance of the capacitance is 1450 Q and 
R p = 20,000 Q. What is the Q of the circuit? 

11.4 In Problem 11.3, the inductance is 250 fiH. What is the frequency of reso- 
nance? 

11.5 At resonance in the circuit in Fig. 1 1.15(a), / = 10 mA and V - 7.5 V. What 
is the resonant resistance of the circuit ? 

11.6 In Problem 11.5 the inductive reactance at resonance is 455 Q. 2nf L; 
what is the value of Q ? 

11.7 Capacitor C in Fig. 11.15(a) is 350 pF. Using data from Problems 11.5 and 
1 1.6, what is the resonant frequency? 

11.8 The half-power points of the frequency response of a resonant circuit are 
647.5 and 662.5 kHz. 



Problems 



281 




(a) 



(b) 



Figure 11.15 



(a) What is the resonant frequency? 

(b) What is the circuit Q? 

11.9 A parallel-resonant circuit as in Fig. 11.15(a) has Q = 150 and / = 0.6 mA. 
The voltage Vh 100 V. 

(a) Find the reactance of L and C at resonance. 

(b) What power is being supplied to the circuit? 

11.10 Using the circuit in Fig. 11.15(a), the voltage V is 55 V; R p =4500£2, L 
= 5 fiH, and C - 0.001 //F. 

(a) What is the resonant frequency? 

(b) What is the circuit Q1 

(c) What is the current / at resonance? 

11.11 A parallel RLC circuit is resonant at 27 kHz. The circuit contains a 0.015-H 
inductance, a 0.002316-//F capacitor, and a parallel resistance of 40,000 Q. 

(a) What is the circuit impedance at resonance? 

(b) What is the circuit Qt 

(c) What is the bandwidth? 

(d) What is the circuit impedance at / ; ? 

11.12 The circuit in Fig. 11.15(b) is to tune the broadcast band from 550 to 1600 
kHz. The capacitor is varied and has a maximum value of 375 pF. What is 
the value of inductance used, and what should be the minimum value of C? 

11.13 A circuit is resonant at 455 kHz and has a 10-kHz bandwidth. The reactance 
of the inductance is 1250 Q; what is R p of the circuit? 

1 1.14 A parallel-resonant circuit is resonant at 20 MHz, the Q is 1 50, and the react- 
ances are each 750 Q. What is the value of R p and of an R, in series with L 
that might replace R p l 

11.15 In a parallel-tuned circuit, the resistance in series with the inductor is 12 Q 
and the inductive reactance is 1450 Q at resonance. Find the Q of the circuit. 

11.16 The resonant frequency of a parallel RLC circuit is 7.3 MHz. If the Q is 80, 
find the half-power frequencies. 

11.17 A parallel RLC circuit is resonant at 20 MHz. The Q is 200 and the reactances 
are 750 £2. 

(a) What is R p 1 

(b) What resistor should be paralleled with R p lo bring the Q down to 50? 

(c) What are the bandwidths, with and without this resistor? 



282 



Tuned and Video Amplifiers 



11.18 In a resonant circuit, C = 60 pF and L is 1 30 ^H. The Q of the circuit is 150 
and R p = 100,000 Q. What is the bandwidth in hertz? 

11.19 A parallel circuit is resonant at 2 MHz. In order to have the circuit resonate 
at 10 MHz, what must be the ratio of the new capacitance to the original 
capacitance? 

11.20 The circuit in Fig. 1 1.3(a) has L = 0.01 H, C = I ^F, and R, = 3 Q. Deter- 
mine /„, BW, and Q. 

11.21 Two circuits of the type of Problem 1 1.18 are critically coupled. Determine k c . 

11.22 In a double-tuned circuit, k - 2k c and Q - 80. If/, = 10.7 MHz, 

(a) What are the frequencies of peak response? 

(b) What is the circuit bandwidth if the gain is down 3 dB at the dip at/,? 

11.23 A double-tuned transformer has L, = L 2 = 100 [M, Q, = Q 2 = 100, with 
coupling k at 150 percent of critical. The source hasg m = 4500 //mhos. 

(a) Find the values for C, and C 2 for resonance at 1.59 MHz. 

(b) Compute the frequency separation of/ and/. 

(c) What is the voltage gain at resonance? 

11.24 An FET amplifier is to have a gain of 20, extending to the highest possible 
frequency. Find that frequency when shunt-compensated, with q = 0.41. 
Also find L needed if g m = 0.0025 mho, C„ - I pF, C gi = 2.5 pF, wiring 
capacitance = 3 pF. 

11.25 A shunt-peaked video amplifier uses an FET with g m 0.0025 mho, C, d 
= 2 pF, C„ - 1.5 pF, C» = 2 pF, and R, = 4000 fi. When designed for 
q = 0.41 find the/; frequency and the needed value of L. What is the low- 
frequency gain? 

11.26 A pulse chain of0.5-//s pulses is sent at a rate of 4000 per second. What ampli- 
fier bandwidth will be needed for reasonable pulse wave form at the output? 
For excellent wave form ? 



12 

Power Amplifiers 



We now wanl to put our signals to work, and large power outputs are required 
for such purposes as driving loudspeakers or scrvomechanisms or transmit- 
ting radio signals through space. Amplifiers providing large power outputs 
require large input signals. The small-signal equivalent circuit cannot be 
used and we must go back to the graphical method of transistor or tube 
analysis. Since the output curves are not linear for large-signal excursions, 
we have to determine the output wave distortion and devise circuits for 
reducing that distortion. 

For maximum power output the loads should match the output resis- 
tances of the amplifiers. Direct matching of output resistance and load does 
not always happen. More usually we have to use an impedance-transforming 
circuit or transformer to supply the matched load conditions necessary for 
maximum power transfer. 

Therefore, in our analysis of power amplifiers we shall be interested in 
such items as distortion levels, power efficiency of the transistor, elimination 
of the heat losses, and the impedance-matching circuits. 

12. 1 Defined Operating Conditions 

Three broad areas of operating conditions are defined for power amplifiers, 
dependent on the chosen bias and the input voltage amplitude. We use the 
letters A, B, and C to designate these conditions, as demonstrated for a tran- 
sistor in Fig. 12.1. 

283 



284 



Power Amplifiers 





Class A 



Class B 



(a) 



<l» 




Class C 



(O 



Figure 12. 1 Transistor operating conditions. 



For Class A operation, the bias is .selected to place the Q point near the 
center of the transfer curve relating v Be and i c . Collector current is present 
for all values of signal input, as shown. This is the operating condition of 
the small-signal amplifiers in our preceding discussions. Distortion is low 
but the power output is also low because of the small input signal. The effi- 
ciency of conversion of dc power to ac power is limited to 50 per cent. Prac- 
tical transistor amplifiers can reach 45 per cent. 

With Class B conditions, the bias is selected to place the Q point on the 
cutoff line and collector current is present only on the forward half cycle 
of the input voltage. During the reverse half cycle, collector currcnf is not 
present in the output of the transistor. Distortion is therefore high but the 
efficiency of power conversion can reach a theoretical maximum of 78.5 per 
cent. Push-pull circuits are employed to supply the missing half cycle and to 
reduce the distortion to usable levels. Practical efficiencies reach 65 per cent. 

For operation in Class C, the bias is set to two or more times the cutoff 
level and current occurs only in short pulses near the forward peak of the 
input wave. Theoretical efficiency reaches 100 per cent but the distortion is 
so high as to limit the circuit's application to radio frequencies where the 
distortion harmonics can be filtered out by resonant circuits. Practical 
efficiencies reach 80 per cent. 

Transistor power amplifiers employ the C-E circuit because of its high 
power gain; similarly, tube power amplifiers use the grounded-cathode cir- 



The Ideal Transformer 



285 



cuil. Class A and B conditions yield amplifiers suited to audio-frequency 
power amplification and Class B and C amplifiers are used at radio frequen- 
cies. Because of the special methods required for the analysis of Class C 
amplifiers with current pulses of varying length and because of their limited 
application, the Class C amplifier svill not be discussed here. 

12.2 The Ideal Transformer 

The transformer in Fig. 1 2.2, with a laminated iron magnetic core, is often 
used as an impedance-transforming device in amplifiers operating at audio 
frequencies from 50 to 1 5,000 Hz. 

Iron Core 




Figure 12.2 The ideal transformer. 

The transformer is efficient in handling power. The voltages and currents 
present in the primary and secondary are related by the turns ratio 



a — 






giving the voltage and current ratios as 

The secondary load is 

On the primary side of the transformer we have 

But from Eq. 12.2 we can write 






*,=$ 



ii 



*' -77 



(12.1) 

(12.2) 
(12.3) 

(12.4) 



and inserting these values into Eq. 12.4 we have 

R . .eV % _ .V* 
R >-7Ja- a h 



286 



Power Amplifiers 



But K 2 // 2 - R, by Eq. 12.3 so 

*i = « 2 *: (12.5) 

A load in the secondary appears in the primary as a resistance a 2 R 2 , with ac 
voltage applied. We say "appears" because /?, is present only when alter- 
nating voltages are present; a direct current does not affect a transformer. 

Thus a loudspeaker of 4 ft can be made to appear as 400 ft on the pri- 
mary side if we use a transformer with the turns ratio 

fl =, yi|° _ yroo - 10 

Such an impedance-transforming device is called an ideal transformer. 
Actual transformers closely approach the ideal transformer performance. 
The transformer also isolates the dc component of collector current from the 
load. For dc the transformer primary appears as the low dc resistance of 
the primary winding, often considered to be zero. 



12.3 Power Relations in the Class A 

Amplifier 

For the large-signal amplifier with an output-matching transformer and 
under Class A conditions, the circuit is that in Fig. 1 2.3. Calculation of 
performance must be derived from the 'graphical output characteristics 
because of the nonlinearity of the transistor curves for large-signal excur- 
sions. 

To obtain large power output we operate the transistor with large power 
input and the internally developed heat is the limiting factor. The input 
power is obtained from ihe collector supply and partially converted to signal 







Figure 12.3 The C-E power amplifier. 



power Relations in Ihe Class A Amplifier 



287 



(12.6) 



output power and delivered to R„ through the transformer. We have 
power input =» ac output | losses 

YcJc = Wi + {VceIc + /^dc) 
where R ic is the dc resistance in the collector circuit, outside the transistor, 
V CB l c being the steady transistor loss at the Q point. Equation 12.6 shows 
that the transistor loss or dissipation is 

P* - VceIc = V cc l c - l c R ie - J 2 R, (12.7) 

The signal output term is /*/?, and as it increases, the transistor loss K C£ / C 
must decrease and the transistor operates cooler in Class A. 

We must design for the worst possible case of no signal, however, so we 
have the £-point loss as 

max />< = V cc l c - I c R ie (12.8) 

The primary winding of the transformer usually has negligible dc resistance 
and in most amplifiers 

ft* = Ra (12.9) 

as the value of the emitter bias resistor. Resistance R E is also kept small, 
however, since its resistance reduces the overall power efficiency and we 
shall neglect it in this somewhat idealized case. Therefore, 

max P d - Vcclc - I C R E = V cc l c (12.10) 

and this establishes the expected transistor loss in the circuit. 

The limiting hyperbola in Fig. 12.4 is drawn for an allowable transistor 
loss at a specified operating temperature; it may be less than the indicated 



AC Load 
Line 




"mm 



Figure 12.4 Construction of the ac load line. 



Power Amplifiers 

loss from Eq. 12.10 and selection of that allowable loss is a thermal problem 
that will be discussed shortly. The Q point will be located on or below the 
limiting dissipation curve. 

The presence of the transformer introduces a new step in the construc- 
tion of the load line since with a transformer the dc and ac loads are not the 
same in resistance. We first select a V cc , usually one-half the maximum 
rated collector-emitter voltage V CElmtx) . We then draw the dc load line, which 
always starts at V cc . Since the dc resistance of the collector circuit in Fig 
12.3 is assumed zero, the slope will be - l//? de = -1/0, and the dc load line 
will be vertical as shown dashed in Fig. 12.4. The intersection with the locus 
will establish the Q point. 

The presence of R E would have given the dc load a slight slope to the left 
but would not alter the general situation. This is discussed in the example 
that follows. 

Draw the ac load line from 2V CC on the x axis and through the Q point. 
The y intercept of the ac load line establishes / mtI and the slope of the line is 
that of R„ which is the primary load to be supplied by the transformer. That 
is, from the slope 



/?,= 



2V, 



cc _ Ycc 



lc 



(12.11) 



since / ra „ = 2/ c by the geometry of the figure. The power loss is P t — V rc l c 
and we can also write 






The base current will be driven symmetrically to some values at I B , and 
I B , as shown, and the transistor will reach peak collector currents as i mtx 
and /„,,„. The power output is found from these peak values of a sine wave as 






~~ 8 



(12.13) 

From the figure / c - (/„,„ - i m J/2 and the efficiency of conversion of 
the dc power to signal power is 



eff. - £ x 100% - 



{'mil <mln)°l 



4r, 



cc 



X 100% 



(12.14) 



The efficiency depends on the amplitude of the output signal. For the greatest 
possible output signal that swings over the complete length of the load line, 
we have / ra!n = and 2y cc /R, = i mp Then 



eff. = 



-0 



= 4^ = 50% 



4K CC //?, " 2i m „ 

and this is the maximum theoretical power conversion efficiency for the 
transformer-coupled Class A amplifier. 



(12.12) 

I B , and 
as / 



Power Relations in the Class A Amplifier 



289 



The saturation voltage limits /„,, to values less than the theoretical but 
because of the low saturation voltage of most transistors a practical amplifier 
can approach the theoretical figure. Vacuum tubes, with much higher satura- 
tion voltages, rarely exceed 30 per cent in conversion efficiency. 

Example: Select a load and determine the power output and conversion 
efficiency for a signal i„ == ±60 mA, maximum possible, applied to the tran- 
sistor in Fig. 12.5. It is rated Fcflfiux) = 65 v > p * aB 30 w un der the expected 
operating conditions and with R B = 1 Q. 

We first draw the maximum-power dissipation curve for 30 W as shown 
and select 

V cc = f^£ = ** = 32.5 V 

A bias resistor R E = I £2 and so the dc load line is drawn with a — 1/1 slope 
to 1 A and 32.5 - I = 31.5 V. The Q point is then found at I c = 0.94 A. 
The ac load line is drawn from 2V CC — 65 V through the Q point and ^, c 
is found from the slope as 

R — ^ CE — — 34 Q 

but R, = R, e — R E — 34 — 1 = 33 £2. We need a transformer to transform 




20 Vcc 4° fi0 

v CE (V) 

Figure 12.5 AC load line. 



290 Power Amplifiers 

R, = 4 Q of a loudspeaker to 33 €1. The turns ratio must be 

«-Vf-2. 8 7 

With the designated input base signal we swing along the ac load line up 
to i„ m> 120 m A and down to i B = 0, giving / m „ = 1.70 A and f„„ = 0.05 A. 
The power output is 

, t JLW-Myx33 ta „ 2W 

The power input to the circuit is 

/>,„ = 32.5 x 0.94 = 30 W 
as expected and the power conversion efficiency is 

efT. -^x 100 = 37% 



12.4 Voltage Limitations 

The manufacturer specifies a maximum safe value for V cz but the transistor 
has some physical limitations on the voltage that may be applied. 

The collector-base depletion layer widens as the collector voltage is 
increased and the depletion layer may extend completely through the thin base 
region at some high voltage. This is known -as punch-through. The transistor 
under this condition appears to have a short circuit between emitter and 
collector. The base loses control until the collector voltage is reduced. 

Avalanching of charges may occur in the collector region at a high voltage; 
however, no physical damage results. This is known as first breakdown. If 
the avalanching current channels into small areas, a hole may be melted 
through the base. This is known as second breakdown and the transistor is 
destroyed. 

The maximum voltage for vacuum tubes is determined by the insulation 
limits and is specified by the manufacturer. 



72.5 Effect of the Thermal Environment 

In order to keep the collector-base junction leakage current small in compari- 
son to the signal current in the collector, it is necessary to keep the temperature 
of the junction below 200°C (392°F) for silicon transistors and I00°C (2I2°F) 
for germanium transistors. 

The rate of heat removal from the collector junction is proportional to 
the difference in temperature between the collector and the ambient sur- 



Etfecl otthe Thermal Environment 



291 






roundings of the transistor. With 7', as the junction temperature and T A as 
the ambient temperature in °C, 

T 3 T A = d ]A P d (12.15) 

where JA is the thermal resistance of the transistor case and mounting in units 
of °C/W. The thermal resistance really states the temperature differential 
needed per watt of heat removed from the transistor. 

The power level for which the dissipation curve of a transistor is drawn 
is dependent on the thermal resistance of the transistor case and its mounting 
and on the ambient temperature surrounding the equipment. In small-signal 
amplifiers, P d is usually at milliwatt level and air convection and conduction 
by the transistor leads is sufficient to remove the generated heat. For higher- 
power amplifiers the transistor mounting is designed to dissipate the heat 
by metallic conduction and convection by the air or by forced-air cooling. 

The collector junction is usually in good thermal contact with the case. 
Electrical insulation is obtained by mounting on a base plate with a thin mica 
insulator with the air pockets filled with silicone grease. The base plate may 
be fitted with convection fins and is then known as a heat sink. This improves 
the heat transfer to the air. But even the heat sink is not able to reduce the 
case temperature to that of the surrounding air and the junction temperature 
will be above that of the case. The air may be considerably above 25°C as 
well. The characteristics of a small heat sink arc given in Fig. 12.6, with 
Oca = 3°C/W. 

Suppose that we have a transistor in which 6-VV dissipation produces the 
junction limit temperature when the case is Held at 25°C in a water bath. But 
in air at 25°C the case temperature will be much above 25°C because of the 



Aluminum Fins Transistor 

,n n n n n/n n n 



.§ 60 
.c 

£ 
< 



o 

-C. 



4(1 



-r\— 

BE 
-10 cm 



1 2.5 cm S 
r cc 



1 20 



o. 

E 



f- 



10 20 30 

Watts in Free Air 

(a) m 

Figure 12.6 (a) Finned heal sink and transistor, MS-10; (b) temperature rise, 
thermal resistance ■> 3°C/W. 



292 



Power Amplifiers 



poor heat transfer from case to air. The junction temperature must rise by 
the amount the case temperature exceeds 25°C in order to continue to conduct 
6 W from the junction. Therefore, with a higher temperature for its sur- 
roundings, we must derate the allowable transistor power level. This is con- 
firmed if we write Eq. 1 2. 1 5 as 

Tj = 6 JA P d ~T A (12.16) 

which indicates that the junction seems to float at some constant differential 
above the ambient. For constant power loss, a 5°C rise in ambient will cause 
a 5"C rise injunction temperature. In fact, the increase may be greater because 
as Tj rises, I CBO and h FE will rise, giving a further increase in I c and junction 
temperature. 

The heat removal situation is demonstrated by the electrical analog in 
Fig. 12.7(a), with two resistances in series. The thermal resistance circuit is 

0ja = jc + ca (°C/W) (12.17) 

where 

6 JA = total thermal resistance, junction to ambient 

6j C — junction to case thermal resistance (supplied by the manufacturer) 

Oca — case to ambient resistance of mounting 

The manufacturer supplies a value for 0, c as well as a thermal derating curve, 
as shown in Fig. 12.7(b) for a transistor with 30- W dissipation at 25°C case 
temperature. From the slope of the curve above 25°C we can find that 



0jc = T J<U rt-T c 



(12.18) 



For this transistor we have T, = 200°C, T c = 25°C, and P t = 30 W so that 
Ojc = 6°C/W. 

40 



Transi^or 




30 



°^ 20 

5 



Ileal Sink 



nnr 



Ambient 



10 





























































J 


lope 


= 0j 







































































50 100 ISO 200 

Case Temperature (°C) 

(a) (b) 

Fixure 12.7 (a) Analog of the thermal circuit ;(b) derating curve for a transistor, 
30 W at 25°C (77°F) or below. 



Determination of Output Distortion 



293 



Substitution of Eq. 12.17 in 12.15 gives us 

Oca d Vjc 



(12.19) 



which determines the allowable thermal resistance for the transistor mounting 
and heat sink. 

The use of the derating curve can be shown in the following examples. 

Example 1: What power rating can we assign to the transistor in Fig. 
1 2.7(b) when used in a mounting and heat sink having Q CA = 3°C/W and with 
Ta - 45°C? 
We have 

Oja — Ojc I Oca 

= 6 + 3 = 9°C/W 
From Eq. 12.15, 



- Tj-Ta 200 - 45 155 

= 17.2W 



and this establishes a value for the limit hyperbola for this transistor under 
the specified temperature conditions. 

The case temperature at 17.2 W can be found from the derating curve 
as 100°C. 

Further data indicate that this transistor (RCA 40316) has a thermal 
resistance when operating in free air of 30°C/W and a maximum dissipation 
in that condition of 6 W. Thus the value of JA = 9°C/W for the mounting 
system provides a considerable improvement in power rating over the free-air 
condition. 

Example 2: Consider the transistor (in the previous example) operating 
at an ambient temperature of 35°C with P d «■ 20 W. What is the maximum 
allowable value for Oca" 1 - 

By Eq. 12.19, 

Oca - 20 ° 2 q 35 - 8.3°C/W 
From the curve the case temperature will be 82°C. 



12.6 Determination of Output 

Distortion 

Distortion of the output waveform occurs by reason of the nonlinear nature 
of the characteristics of the transistors employed. A current transfer curve 
is plotted in Fig. 12.8, resulting from points chosen along the load line for 
the amplifier in Fig. 12.5. For large-signal excursions we would anticipate 



294 



Power Amplilicis 







H cos oil 



Figure 12.8 Current transfer curve for the load line of Fig. 12.5. 



some distortion of the i c waveform, or flattening of positive and negative 
peaks with the Q point shown. 

The distorted waveform can be described by a fundamental frequency 
sinusoidal wave added to various amplitudes of harmonically related sine 
waves. The harmonics are at integer multiples of the fundamental frequency. 
The wave may be studied by use of a wave analyzer, which provides an 
amplitude reading for each harmonic frequency present. 

We can also analyze amplifier outputs by reading amplitudes from the 
transfer curve, however, using i c values resulting from an assumed sinusoidal 
base current signal, as shown in Fig. 12.8. We write the collector current 
wave as a Fourier series of cosines: 



ic - A„ 



(12.20) 



I A x cos cot -'- A z cos 2cot + ■•■ 

The amplitudes of the several frequencies present can be determined by 
evaluation of the A ,A u A z ,... coefficients. The transfer curve can be used 
to find i c at three points in time, spaced over the positive and negative halves 
of the input base current cosine wave. Substitution of these values into Eq. 



Determination of Output Distortion 

12.20 results in three equations: 
At cot - 0, i c «= /„,,, 

imMx = A a + A,cos (0) H A z cos 2(0) 

'mix = A.+ A, - A, 

since cos (0) = I. 

At cot = n/2, i c - I c : 

I c - A. + A, cos (y) I- A z cos 2( T ) 

I C =A - A, 

since cos (a/2) = 0, cos n = — I. 
At cot — 7t, i c = i min : 

'mm *■ A„ + A, cos (n) + A, cos 2(n) 
'mm — A a — A x -r A % 

since cos % «= — 1, cos 2w = I. 

Summarizing, we have 

'mix = A„ + A i + A i 
I c = A a - A 2 

'mm = A„ — A, -)" A 2 

and solving simultaneously 



295 



A % - A, = a 



'mln - 2/ c 



A 'mix " 'mln 

A * 2 

The second-harmonic distortion present is the ratio 

% D z = 4 1 X 100 



(12.21) 



(12.22) 



(12.23) 



(12.24) 
(12.25) 

(12.26) 



Since A, is the peak value of the fundamental frequency, the fundamental 
output power is 

*»-(4fc)'*»-TM« (,2 - 27) 

Had we used five points along the input current wave, adding two currents 
as i x and /, at half amplitude at cot = n/3 and cot = 2b/3, we could calculate 
additional harmonics. The equations are 

A. = $(/.„ : '„,,„) + H'« + /,) - Ic 02.28) 

^ - -J(' m „ - /.,.) + & - Q ( 12 - 29 ) 

A - *('m.x + /«,.) - i/ c (12-30) 

A S ~ |CU - '-!.) - & - y (12-31) 

^ = ^(/m., + /„,.) - K<", + y + i'c (12-32) 



296 



With 



Z) 2 = ^i x 100%, 



z> 3 = 4*x 100%, 



Power Amplifiers 



04 = 4** 100%, 



the total-harmonic distortion is 



D = JVi+D< s + Di+ ... (12.33) 

It is convenient to be able to recognize harmonic content from the wave- 
form and so we present Fig. 12.9. The waveform in Fig. 12.9(a) in which the 
positive and negative waves are not similar contains predominantly even- 
order harmonics, whereas that in Fig. 12.9(b) with the two halves as mirror 
images contains odd-order harmonics. 

Levels of allowable distortion are subject to individual judgment but a 
total distortion of 5 per cent is usually tolerated. The allowable level is reduced 
to less than 1 percent when high-fidelity equipment is involved. In most power 
amplifiers this low level of distortion can only be achieved through use of 
negative feedback. 





< a > (b) 

Figure 12.9 (a) Even harmonics present ; (b) odd harmonics present. 

Example: The values for waveform analysis can be read from Fig. 12.8 as 

'm.x=l-70 I. = 1.40 







/ aln = 0.05 


i y - 0.55 






I c = 0.94 


We have 












'mix 'rnln 


- 1.65 






'ma* ~f* 'mln 


= 1.75 






h - i, 


= 0.85 






I* + i, 


= 1.95 


Using Eq. 12.28 to 12 


.32 


we have 





A. = ^ + i^ - 0.94 = 0.002 



The Push-Pull Circuit and Class B Operation 

1.65 . 0.85 
3 

1.75 
4 

1.65 0.85 



297 



A, = *f- + ^ = 0.833 



A, = ^- 0.47 = -0.033 



A,= 



-0.0083 

-0.034 



3_ ~6 5 

. _ 1.75 1.95 . 0.94 

The distortion is predominantly even order, due to the second and fourth 
harmonics. We find 

^3 = 5^x100=1% 



D« = 



0.833 
0.034 



x 100-4.1% 



Total distortion : 



0.833 
D = V4M- P + 4.1 2 = 5.7% 

12.7 The Push-Pull Circuit and Class B 

Operation 

If we move the Q point down the transfer curve toward the origin, our for- 
ward swing can still drive the transistor up to i' miX . Therefore we have a 
greater positive i c current swing and greater power output. The other half of 
the input wave drives the transistor toward cutoff, however, and we have an 
unsymmetrical positive and negative wave, as in Fig. 12. 10. While achieving 
greater power output from a given transistor, we have generated greater even- 
harmonic distortion. 

The bias conditions as shown in Fig. 12.10 place the transistors at cutoff 
for part of a half cycle, the operating condition is intermediate between Class 
A and Class B, and it is known as Class AB. 

We can cancel most of the distortion if we add a second transistor Q lt 
oppositely connected as in Fig. 12.11(a), and achieve the power output of 
two transistors. The input transformer 7", supplies opposite polarity signal 
voltages to the two bases, with respect to the midpoint of the winding. On 
the first half cycle, transistor Q K is driven in the forward direction by the input 
signal and transistor Q 2 is driven toward cutoff by the opposite polarity 
input voltage from the other half of the winding. On the next half cycle, 
the input voltages reverse and Q, is driven toward cutoffand Q 2 gives a full 
forward output. These output currents combine in the output transformer 
7" 2 ; as the transistors appear to operate oppositely on each half cycle, the 
circuit is called a push-pull amplifier. 









298 




Figure 12.10 Class AB operation. 

The circuit action is similar to that of a differential input-differential out- 
put amplifier; had in-phase signals been applied to the two inputs, the net 
output signal would be zero. The opposite-phase input signals provide an 
output after being subtracted in the output transformer, however; that is, 
+^ — (— A) = 2A. The even-harmonic components are generated in phase 
and cancel in the output. Figure 12.12 illustrates how the fundamental waves 
add and the second harmonics cancel upon subtraction in the transformer. 

Accuracy of harmonic cancellation is assured only with balanced transistor 
parameters and equal input voltages of 180° phase relation. 

If we move the Q point to cutoff on the transfer curve, we have com- 
plete cutoff of one-half of the input wave and can show the action of the 
second transistor by an opposite transfer curve drawn to the common origin, 
as in Fig. 12.13. With cutoff bias, the operating condition is Class B. Each 
transistor supplies an independent half wave and these are combined in the 
output. Use of the push-pull connection cancels the inherent even-order dis- 
tortion of the Class B operation and allows us to utilize the higher efficiency 
and power output of Class B. 

The current components i Ct and / c , in the output transformer are 



fc. = h sin cot + J c , 
ic, = —I a sin cot + I Ci 



(12.34) 
(12.35) 



where /„ sin cot and -/. sin cot are the oppositely phased ac signals and I c , 
and I c , are the steady 0-point currents. Passing in the windings of transfor- 













?>'cc 




<y 


V 


<3 




<4 


C 


9 < 


:*■ 




€ 
o 
O 


5 i 


K i 




c 


» 


\ AAA* , 




o 


> 
o 

I 


VVVV " 


.1 


G 

a 
c 
o 


» 






c 


■ 






c 


» 






°f 


» 


.•"""""J 




c 


» 


rk*F 


V 


c 


» — 


(a 


)ft 


o 




\ k 


<: 




\b> 


/ 


o 



T, 




(u) 




#4* 



*, 



i 






<c, 




(b) 

Figure 12.11 The push-pull amplifier. 




Fundamental 
» - Second 
\ / v^ Harmonic 

|i / > 
rt-T •— r 



«2i> 




Fundamental 




Second Harmonic 



fii'urc 12.12 Waveforms in the push-pull output. 



299 



300 



Power Amplifiers 






Output Q. 




Figure 12.13 The action of two transistors under Class B conditions. 

mer 7" 2 in opposite dircclions, these currents are subtractive in their effect 
on the secondary load so that if I Cl = (— )/ Cl , then 

ii = K('c, ~ W = 2KI„ sin cot (12.36) 

The signal currents appear added in the secondary and any dc components 
are canceled. Thus the steady magnetomotive force that might saturate the 
iron core is removed and the transformer core is more effectively used. 

Since the g-point current is essentially zero in a Class B amplifier, the 
power supply is called upon for sudden current surges as the signal input 
varies. Power supplies for Class B circuits should maintain constant voltage 
through these surges or have good voltage regulation. The use of shunt-C 
filters is advisable for such applications. 

Since the dc input power is low for zero and small signals, Class B 
amplifiers arc also preferred for battery-operated equipment. 



12.8 Performance of a Class B Push- Pull 

Amplifier 

For study of a Class B push-pull amplifier we have the output characteristics 
for Q, as drawn in the upper half of Fig. 12.14, and with Q 2 in a subtractive 
relation to Q t in that its characteristics are drawn upside down. The factor 
common to both transistors is V cc , and the v CE axes are aligned at that 



Performance of a Class B Push-Pull Amplifier 



301 



voltage. Since the transistors operate with zero / c at cutoff, the V cc point at 
which the curves are aligned is also the Q point of the amplifier. Signal 
swings occur up and down the load line from that quiescent point. The ac 
load line shown represents the largest output swing, from the upper knee of 
Q, to the same knee position for Q z . The slope of the load line represents the 
primary transformer load R,, presented to each transistor, as 



*, = ££ 



(12.37) 



We have assumed the saturation voltage to be small, with a current /„. 

On the first half cycle, the input signal voltage is forward to Q t and drives 
the operating point up the load line toward /„; this represents operation below 
cutoff for Q 2 . On the next half cycle the input voltage is forward to Q 2 and its 
operating point is driven along the load line toward its /„ value; this is below 
cutoff for Q x . 




Figure 12.14 The Class B composite load line. 

The output current consists of two separate half sine waves, combined into 
a full sine wave by the output transformer, and gives the output waveform 
shown in Fig. 12.12. Using what we have learned about such waveforms in 
the study of diode rectifiers, the dc value of current for one transistor is 

/dc. - - 

where I m is the peak of the sine wave. For both transistors 

/ac - 21*, = ^ (12.38) 

n 



302 

and the dc power input to the amplifier is 

p 2/ M V cc 



Power Amplifiers 



(12.39) 



/ =4t, 



The ac current represented by the composite output wave of Fig. 12.12 is 

(12.40) 
and the ac power output to the transformer is 

P., -- I -fR 1 (12.41) 

The power conversion efficiency of a Class B amplifier is 

eff =fe x,00O /« = 2TO x,00% 

Because of Eq. 12.37, we can write 

eff . = WUAWccll.) x ,00% = «& X 100% (12.42) 

'm'CC 4'o 

It should be possible to drive the peak of the signal output wave to where 
I m = h on the transistor curves. Under the maximum signal producing this 
large peak current we have 

max eff. = -J. x 100 = 78.5% 

as the maximum theoretical conversion efficiency. This is a considerable 
improvement over Class A conditions and is one of the reasons for the wide- 
spread use of Class B amplifiers. 

By use of Eq. 1 2.38 we can write the signal power output as 



P - n 2 HR, 



(12.43) 



which shows that the dc power input increases with increasing input signal. 
The peak transistor dissipation does not occur at maximum output but 
instead occurs at a signal level that is 40 per cent of that maximum. The peak 
power loss for use in transistor selection is 



max/> d = Ai^c=o.20^ 

71 f<| A, 

at which point the conversion efficiency is 50 per cent. 



(12.44) 



12.9 Output Circuits without Transformers 

The output load of most power amplifiers is a loudspeaker and these usually 
have a resistance of 3 to 16 £1 At this low level of resistance there is no need 
for impedance matching and loudspeakers can be directly employed as loads 



Output Circuits without Transformers 



303 



for power transistors. Thus the output transformer, with its problems of 
weight, size, cost, and frequency distortion, can be eliminated in appropriate 
push-pull circuits which cancel the dc currents which might otherwise appear 
in the speaker circuits. 

The usual circuits are derived from the bridge circuit shown in Fig. 1 2. 1 5(a), 
in which two push-pull circuits drive a common load 7?, between y and 
z. These circuits are, respectively, Q, and Q 3 as the upper amplifier and Q 1 
and Q t as the lower amplifier. The four input windings have a common 
primary, not shown, and the dots indicate the simultaneously positive ter- 
minals. Transistors Q l and Q, are simultaneously driven upward in current 
and Q 2 and Q } are driven downward at the same time. Thus point;' is raised 
in potential and z is lowered; on the next half cycle y goes down and z goes 
up in potential so that we have an ac voltage across the load. Points y and z 




(b) to 

Figure 12.15 Development of circuits without the output transformer. 






304 



Power Amplifiers 



appear to teeter-totter in voltage around point x as a fulcrum. With no 
signal, points y and z are in dc balance and no dc current passes through 
the load. 

The circuit in Fig. 12. 1 5(b) is formed by splitting the first circuit along the 
common or ground line at g, g; transistors Q, and & are eliminated. Point 
y moves up and down as Q, and Q, are driven forward and reversed and an 
ac signal is present. Point x is maintained at its previous dc potential by 
splitting the power supply into two sections. The value for R L is one-half of 
load resistance /?,. 

The circuit in Fig. 12.15(c) evolved to avoid the expense of two power 
supplies. Since point .v was at zero signal potential to ground in Fig. 12. 1 5(b), 
it can be connected to actual ground if the large blocking capacitance C is 
used to avoid short-circuiting the dc supply. Capacitance C is usually of 
several thousand microfarads so that its reactance will be small compared 
to the resistance R L of 3 to 16 SI. This R L C combination establishes a low- 
frequency limit for the amplifier as 

f - l 
Jx ~2nR L C 



12. 10 Phase Inverters for Push- Pull Input 

Two equal voltages at 180° in phase are needed for push-pull amplifier input. 
The size and cost of input transformers can be avoided by use of circuits 
known as phase inverters; examples are shown in Fig. 12.16. 

The circuit in Fig. 12.16(a) is called a phase splitter; it consists of an in- 






(a) 



(b) 
Figure 12.16 Phase inverter circuiis. 



.' 




Complementary Symmetry Circuits 



305 



phase output across R E at V n and a reversed-phasc output across R at V 0l . 
In reality the circuit is an emitter follower with a collector load added. Choos- 
ing R and R E to provide a gain of unity to V 0l , we have balanced voltages 
since the emitter follower gain is also near unity. 

A second phase inverter is that in Fig. 12.16(b), which consists of the 
preceding circuit with Q 2 added. Both outputs are now taken from low- 
output resistance emitter followers and this is better for providing balanced 
and good waveform signals for Class B amplifiers, where there may be 
demands for large currents to drive the power transistors. 

A differential amplifier may also be used as a phase inverter to drive a 
push-pull output stage. 

12.11 Complementary Symmetry Circuits 

The requirement for equal and oppositely phased input voltages for push- 
pull amplifiers is eliminated by use of a matched pair, npn and pnp, com- 
plementary transistors. As shown in Fig. 12.18(a), a signal of positive polarity 
to ground will simultaneously drive the npn unit Q l into forward conduction 
and the pnp unit Q 2 into cutoff. Thus a common input voltage will give Class 
B operation of the push-pull circuit. Figure 12.1 8(a) also employs a Darlington 
compound connection for higher input resistance. The Darlington transistors 
are also of complementary form. The push-pull circuit is basically that shown 
in Fig. 12.15(b), with R, as the load. 

Figure 12.18(b) shows a driver stage using Q t and Q 2 as complementary 
symmetry transistors with a common input voltage. Their outputs, as emitter 
followers, provide low-resistance and high-current sources for normal pnp 
transistors Q 3 and Q 4 in a Class B push-pull circuit whose basic circuit form 
is that in Fig. 12.15(c). 

Crossover distortion, as in Fig. 12.17, can appear in the output currents 
of such amplifiers because of the lack of symmetry in ihc pnp and npn char- 
acteristics near cutoff. One unit will not turn off at exactly the same currents 
and voltages as the other turns on. A similar phenomenon may be encoun- 
tered in Class B amplifiers using the same transistor types and is corrected by 
bias adjustment and use of ample negative feedback. 



v 



w 



Figure 12.17 Crossover distortion in a waveform. 




(b) 
Figure 12.18 (a) Complementary symmetry power amplifier; (b)complementary 
symmetry driver for a Class B push-pull amplifier. 



306 



The Class B Linear Radio-Frequency Amplifier 



307 

12. 12 The Class B Linear 
Radio-Frequency Amplifier 



A Class B amplifier, with its accompanying high-power efficiency, may be 
operated single-ended or push-pull when used at radio frequencies with a 
resonant load. Because of the near linearity of the transfer curve between 
input ;' B and output i c , the output voltage is proportional to the driving 
voltage. The title of Class B linear amplifier is intended to emphasize this 
point as the amplifier is used to develop radio-frequency power when driven 
by varying amplitude or modulated radio-frequency voltages. 

Since the resonant frequency of the tuned load circuit is high at f , in 
Fig. 1 2. 19, the distortion components generated will have even higher fre- 
quencies at 2f„, 3f a Having frequencies of two or more times the tuned 

frequency of the circuit, the harmonics can be well filtered out by a resonant 
load circuit. To discriminate against these harmonics the load circuit Q is 
usually maintained in the range from 10 to 15. 

To retain its property of linearity, the transistor should not be overdriven 

Resonant lo/„ 




(■> 




(b) 

Figure 12.19 (a) A Class B linear radio-frequency amplifier with tuned load; 
(b) the n-matching network. 



308 



Power Amplifiers 



since that will take it into the flattened saturation portion of the transfer 
curve above /„„ in Fig. 12.8. 

A common modification of the resonant circuit, the n-mat citing network 
in Fig. 12.19(b), provides increased filtering action for the undesirable har- 
monics. The output power is normally supplied to an "antenna of 50 to 75 Q 
as R 2 and the load facing the transistor is R, of several hundred ohms. With 
Q greater than 10, the circuit elements of then network should be 






S! X c 



-Si 



R,R, 



Q 1 - RJR t 

Y — $.1.(^1 ^*\ 



(12.45) 
(12.46) 



Since the reactances of the shunt capacitances decrease with higher fre- 
quencies, they provide low impedance paths which short-circuit the higher- 
order harmonics to ground. This is why the n circuit is preferred to the T 
circuit. 



12.13 Summary 

The power output from transistors and tubes is limited by the internal losses 
that must be removed as heat. Only cooling by conduction or convection is 
possible at the low operating temperatures of transistors, although heat may 
be removed by radiation at the temperatures encountered with large-power 
vacuum tubes. The internal collector loss of the transistor must be removed 
through the transistor mounting and this is a point needing careful thermal 
design. Heat sinks are available to improve the heat transfer at temperatures 
only 100°C(212°F) to 200°C (400°F) above the ambient air temperature. 

Large-signal excursions for the transistor create nonlinear distortion and 
harmonic frequencies in the output waveform. Suitable loads are chosen 
for maximum output conditions on the transistor output characteristics and 
the resulting harmonic distortion can be calculated directly from the transfer 
curve, relating i B and i c for a given load. 

Class A operation has relatively poor power efficiency and low power 
output from a given transistor or tube but it has small waveform distortion. 
Class B conditions lead to higher efficiency and greater power output but 
the even-order distortion is very large. Push-pull circuits can reduce the even- 
order distortion and negative feedback can be added to reduce the remaining 
odd-order harmonics. These several circuit modifications reduce the dis- 
tortion of the Class B condition to allowable levels and the Class B amplifier 
is especially valuable in the generation of large-power outputs from small 
transistors or tubes or when the dc power supply is limited in capability, as 
in battery-operated equipment. 






r 



Review Questions 



309 



REVIEW QUESTIONS 

12.1 What is meant by / C eo? 

12.2 Why must wc limit the junction temperature of a transistor? 

12.3 Define Class A operation; where would you establish the Q point? 

12.4 Define Class B operation; where would you establish the Q point? 

12.5 Define Class C operation; at what value would you establish the base bias? 

12.6 Compare the relative amounts of power output and distortion from a given 
transistor in Class A, Class B, and Class C operation. 

12.7 What is the theoretical maximum conversion efficiency for Class A, Class B, 
and Class C amplifiers? 

12.8 How close to the theoretical conversion efficiency can you expect to operate 
a transistor Class A amplifier? A vacuum-tube Class A amplifier? 

12.9 What is meant by the variable a as related to a transformer? 

12.10 A certain transformer is listed as "400 to 3.2 SI." What is meant by this, and 
what turns ratio would you find? 

12.11 What is an ideal transformer? 

12.12 What is meant by the locus of collector loss? 

12.13 How does l CB0 vary with temperature? 

12.14 With fixed base-bias current, what happens to the Q point on a load line when 
h FE increases? 

12.15 Define thermal resistance. 

12.16 How do wc use a transistor loss derating curve? 

12.17 How can we find the thermal resistance of the transistor case from the derat- 
ing curve? 

12.18 How would Osa change if we add an air blower on the heat sink? 

12.19 Why are fins put on a heat sink? 

12.20 Define heat sink. 

12.21 What is second breakdown of a transistor? Is it damaging? 

12.22 What limits the value of anode voltage applied to a triode? 

12.23 What is punch-through in a transistor? 

12.24 Explain the purpose of an output transformer in a Class A amplifier. How 
would you determine the turns ratio? 

12.25 Define total-harmonic distortion percentage. 

12.26 Define second-harmonic distortion. 

12.27 What is the difference in form of a wave having even-order harmonics and 
a wave having odd-order harmonics? 

12.28 How many current points would be needed on a transfer curve to find the 
value of the sixth harmonic? 

12.29 What are the advantages of Class B push-pull operation over Class A push- 
pull operation? 



310 



Power Amplifiers 



12.30 What is the advantage of Class AB operation over Class A operation? 

12.31 What feature of Class B push-pull operation makes it especially valuable for 
battery-operated equipment? 

12.32 What is complementary symmetry in transistor amplifiers? 

12.33 What is crossover distortion? 

12.34 What is the purpose of a phase inverter? 

12.35 Why is the phase inverter in Fig. 12.16(b) superior to that in Fig. 12.16(a)? 

12.36 Why does a Class B linear radio-frequency amplifier use a tuned circuit for 
a load ? 

12.37 What kind of signals can a Class B linear radio-frequency amplifier handle? 

12.38 At what value of output power does maximum collector dissipation occur: 

(a) In a Class A amplifier? 

(b) In a Class B amplifier? 

PROBLEMS 

12.1 What turns ratio is needed for a transformer to couple an 8-fi load to a tran- 
sistor requiring 450 Q ? 

12.2 An amplifier with an output resistance of 425 Q is assembled with a trans- 
former having a turns ratio a = 2.57. What should the secondary load be? 

12.3 A transformer is found mounted on an 8-J2 loudspeaker. The turns ratio is 
measured as 10.5: I. What ac primary resistance will be present? 

12.4 A given transistor type has a dissipation rating of 20 W in a certain heat sink. 
What is the greatest possible power output when one of these transistors is 
used in Class A service? In Class B service? 

12.5 A transformer provides a load of 25 £1 to a transistor with the output curves 





.' 



Problems 



311 



shown in Fig. 12.20(a). The primary has a dc resistance of 5 CI. Find the true 
Q point, power output, and efficiency for V C c m 30 V, collector loss of 25 W, 
and driving signal of ± 15 mA peak at the base. 

12.6 Determine the second-harmonic distortion for the amplifier of Problem 12.5. 

12.7 Draw the transfer i B , i c curve for the load line of Problem 12.5. 

12.8 A transistor is derated according to the curve in Fig. 12.21. What is the thermal 
resistance 6 JC ? 

12.9 The heal sink and mounting for the transistor shown in Fig. 12.21 establishes 
the case temperature at 75°C. What is the power loss in the transistor? 

12.10 A silicon transistor is derated according to the curve in Fig. 12.21. For a case 
temperature of 60°C, what is the allowable power dissipation? 

30 



20 





















\ 


















\ 
















\ 
















\ 
















' 


\ 





10 



50 100 150 200 

Case Temperature (°C) 

Figure 12.21 

12.11 If the ambient air temperature is 35°C, what is the allowable value of Q c .t for 
the transistor of Problem 12.10? 

12.12 The junction temperature of a transistor is 130°C. The dissipation at a case 
temperature of 2S°C is 5 W; at a 25°C ambient air temperature it is 2 W. 
What is the value of 0c? 

12.13 A silicon transistor is rated at a thermal resistance d )C - 0.9°C/W with T J[mix) 
= 160°C. 

(a) Find the allowable power dissipation if the case is maintained at 50°C. 

(b) Find the power that could be dissipated if C a = 2°C/W and the ambient 
air temperature is 35°C. 

12.14 A silicon transistor has T Hmtx) = 180°C and 9 JC - 0.7°C/W. If mounted so 
that Oca = 0.9°C/W, find the power dissipation allowable if the ambient 
temperature is 30°C. 

12.15 A transformer-coupled Class A amplifier drives an 8-fil loudspeaker through 
a transformer having a — 4.3 : 1 . With a power supply of K t .-c = 36 V, the 
amplifier delivers 3 W to the loudspeaker. 

(a) Find the ac voltage across the transformer primary. 

(b) Find the rms value of loudspeaker voltage. 

(c) Find the rms value of the loudspeaker current. 






312 

12.16 

12.17 
12.18 

12.19 



12.20 



12.21 



12.22 



Power Amplifiers 

For the following current measurements from a waveform, find the second- 
harmonic distortion percentage: / m „ = 0.9 A, ; mln - 0.47 A, and I c = 0.65 
A. The load resistance through which this current passes is 95 CI. What is 
the fundamental power output? 

If the (2-point dc current is 0.22 A in Problem 12.15, find the conversion 
efficiency. 

Find the collector dissipation for a Class A operated transistor with V^ = 
30 V, I c = 2.0 A, and ac current of 0.7 A. The load is supplied by a trans- 
former with a = 2.2: 1 ; the secondary load is 10 fi. What is the signal power 
output? 

An amplifier has only second-harmonic distortion. 

(>) If »m.x = 250 mA, i m , n = 5 mA, and I c = 100 mA, find the value of the 

second-harmonic distortion D t . 
(b) If a transformer of a = 4 : I couples a 10-fl load to this amplifier, what is 

the fundamental frequency power output? 

Using the transistor in Fig. 12.20(b) with V cc = 40 V, I B = 60 mA at the Q 
point, with 20 Q given by the transformer load, find 

(a) The fundamental power output with an ac sinusoidal input of ±40 mA 
peak. 

(b) The second- and third-harmonic percentages. 

(c) The conversion efficiency. 

A Class B push-pull amplifier is supplied by V cc = 50 V and the signal swings 
the collector voltage to ».- mln = 10 V. The dc loss in both transistors is 40 W. 

(a) Find the power being delivered to the load. 

(b) Find the conversion efficiency. 

With a transformer load of 30 SI = /?, for each transistor, a Class B push- 
pull amplifier takes 0.75 A from the dc supply with a particular input signal. 

(a) What is the ac power output? 

(b) If the dc supply is at 40 V, what is the transistor power loss and the con- 
version efficiency ? 



.' 






13 

Oscillator Principles 



Oscillator circuits are the generators of our radio frequencies. A primary 
requirement is that the frequency be stable but, under conditions of varying 
supply voltages and varying temperature, sufficient stability is difficult to 
achieve. 

There is a variety of circuits available but fundamentally those to be 
studied here all depend on positive feedback for their operation. 



13.1 Oscillator Feedback Principles 

In the circuit in Fig. 13. 1 the voltage fed back from the amplifier output 
supplies the total input 

and since V„ — —AV„ then 

V,= -APY, (13.1) 

from which 

(1 + Afi)V, - 
If an output is present however, then V, ■£■ and therefore 

1 + Afi = 

Ap=-l (13.2) 



313 



314 



Oscillator Principles 




Figure 13.1 The basic feedback oscillator. 

which was the condition wc found in Chapter 9 that would lead to oscillation 
in a feedback amplifier. 

Here we want the circuit to oscillate and so the expression above states 
two requirements for oscillation to occur: / 

1. That/iy? = -1. 

2. That the net phase shift around the feedback loop be 0° or 360° or 
2/i x 180°, where n is 0, 1,2,3, 

In Fig. 13.1 the amplifier provides its own input and initially the gain 
must be such that \Af}\> 1. An initial turn-on surge or noise voltage 
provides V, at the input and this is amplified to the output. This amplified 
output is fed back to the input as a larger signal. The process is repeated at 
successively greater amplitudes until the output becomes limited at cutoff 
and saturation. By operation to those limits the gain is reduced to an average 
level called for by Eq. 13.2 and a steady level is maintained. 

The frequency of oscillation adjusts itself so that the phase shift require- 
ment is satisfied. 

Limitation of amplitude by cutoff and saturation implies distortion. The 
resultant harmonic frequencies can best be filtered out by use of a resonant 
circuit for the /? network. High Q (high C) there provides better discrimina- 
tion against the harmonic frequencies and also causes the circuit to oscillate 
more precisely at the resonant frequency of the tuned circuit. 

The reactance network in Fig. 13.2(a) provides feedback and 180° of phase 
shift between the output and the input; the FET provides 180° of additional 
phase shift to meet the phase requirement of 360° for oscillation. 

Reactances X,, X lt and X } must be resonant and the frequency of oscilla- 
tion adjusts to make this occur. We know that the resonant frequency will 
make 

X, + X 1 + X, = (13.3) 

For this equation to be true, one or two but not all three of the reactances 
must be negative and capacitive. This shows us the possibility of two basic 



Oscillator Feedback Principles 



315 



r4T— 

s 


- X, — J 


— x 2 — 


* 


• ' 



c 

o 







fc-t 



x, 



(a) (b) 

Figure 13.2 (a) Tuned-circuit feedback; (b) the equivalenl circuit. 
i 





Figure 13.3 (a) Colpitts oscillator without bias circuits; (b) Hartley oscillator 
circuit. 

circuits as in Fig. 13.3, the Hartley oscillator with two inductances (a tapped 
coil) and the Colpitts oscillator with two capacitances. 
For the FET circuit in Fig. 13.2(b) wc have 

A = -g m R 

where R is the resonant resistance of X l in parallel with X, + -*V The 
circuit as drawn also shows us that 

P X> + X, 

from our previous work on feedback amplifiers. From these equations it is 
possible to show that the gain requirement for oscillation leads to 



£-> 



AT, 



(13.4) 



which can be satisfied. The frequency requirement of Eq. 13.3 satisfies the 
phase requirement and the circuit will oscillate. 



316 



Oscillator Principles 



In order that g m be positive, Eq. 13.4 shows that X x and X 2 must be the 
same type of reactance; X 3 must therefore be of the opposite type to satisfy 
Eq. 13.3. 

13.2 The Hartley and Colpitts Oscillators 

The Hartley and Colpitts circuits in Fig. 13.3 are the basic feedback oscillator 
circuits. Analysis of the Colpitis oscillator provides an equation for the 
frequency of oscillation: 

ft = , ~ J —== (13.5) 



2njl 



C t C 1 



C f "at 



c, + c 2 

The first term under the radical is the resonant frequency of the L and 
equivalent C values of the tuned circuit. The second term under the radical 
shows that variation of the transistor parameters can have an effect on the 
frequency of oscillation. The effect is small because h a . < h,,. For design, we 
use 

(13.6) 



/o = 



2*V Z 



C,Q In^LC 



C, +C 2 

We are normally concerned with oscillator frequencies of I MHz or over 
and a variation of 0.01 per cent at 1 MHz represents a shift of only 100 Hz; 
later we shall discuss oscillators in which such a shift is not negligible. 

The gain requirement for oscillation for the Colpitts circuit gives 



hf.^. £r + hJ'o,-^ 



(13.7) 



and, with C 2 = C,, we have no difficulty in meeting this requirement. 

The Hartley oscillator can be similarly analyzed for frequency of oscilla- 
tion. Again a small additive term is dependent on hjlt u and the frequency 
can be slightly affected by transistor parameters. Basically the frequency of 
oscillation is 

where L is the total inductance of the tapped coil. 

The necessary gain requirement makes h,, a function of LJL, ; again we 
have no problem in meeting this requirement since we may make L, = L t . 

13.3 Practical Transistor Oscillators 

The basic circuits in Fig. 13.3 do not include the bias circuits and associated 
blocking and bypass capacitors that are required for normal operation. These 
oscillator circuits are shown complete in Fig. 1 3.4. 






' 



Practical Transistor Oscillators 



317 




(a) (b) 

Figure 13.4 (a) Practical Colpitts oscillator circuit ;(b) Hartley oscillator circuit. 

Resistors /?,, R 2 , and R E provide the bias with initial design values 
setting the Q point a little more toward cutoff than for normal Class A 
operation. Full Class B or Class C biasing cannot be used because we would 
ihcn have zero initial current. This would mean that there would be no way 
for the oscillations to start. Our design values for these bias resistors are 
determined by the methods of Chapter 5. Capacitor C E is selected to have 
a reactance less than R E /\0 at the operating frequency. 

The two blocking capacitors C c are placed to isolate the bias voltages to 
the transistor and should have reactances of only a few hundred ohms. 

The radio-frequency choke, RFC, passes the dc collector current but by 
reason of its inductive reactance it prevents current at oscillator frequency 
from reaching the power supply, lt should have an inductive reactance that is 
much larger than the reactance of a blocking capacitor, or 



lOA'c- 



(13.9) 



The frequency of oscillation in each circuit will be very close to the reso- 
nant frequency of the tuned circuit when stray capacitances of the wiring are 
included. The tuned circuit is sometimes called a tank circuit because it 
serves as a reservoir of radio-frequency energy, part of which is inductively 
coupled to the output as V„. 

While values of g m and h fc have been indicated as minimums to assure that 
oscillations will start, the value of C 2 is usually equal to C, in the Colpitts 
circuit because equal capacitance split-stator tuning capacitors are generally 



318 



Oscillator Principles 



available. In the Hartley circuit the choice of the tap position on L is arbitrary. 
It is usually located so that L z is two to four times greater than L, ; this 
does not overdrive the transistor and results in a lower harmonic content 
in the waveform. 

The Armstrong oscillator circuit is shown in Fig. 13.5(a). If examined 
carefully, it can be seen that it is basically a Hartley oscillator circuit (tapped 
inductor) with only the base circuit tuned. The power supply is introduced in 
series with the collector inductance, and the radio-frequency choke prevents 
high-frequency currents from entering the power supply. Instead, these 
currents are provided with an easy path to the emitter and ground through 
the bypass capacitor C c . 




(b) 

Figure 13.5 (a) Armstrong variation of the Hartley oscillator ; (b) tuncd-collcctor 
oscillalor, lapped-collcctor coil. 



Crystal Control of Frequency 



319 



The circuit in Fig. 13.5(b), is a tuned-collector oscillator, also basically a 
Hartley circuit. In order to raise the Q of the tuned circuit, the collector is 
connected to a lap on the circuit inductance. 



13.4 Crystal Control of Frequency 

Piezoelectric quartz crystals arc of hexagonal form, with atomic plus and 
minus charges arranged in the unit crystal as in Fig. 13.6(a). Being symmet- 
rical, the charges balance and the crystal is electrically neutral. Horizontal 
pressure from the left and right reduces angle tf> and the negative charge (1) 
moves up and the positive charge (2) moves down. This movement destroys 
the symmetry and the crystal shows a negative charge at the top surface and 
a positive charge on the bottom surface. Horizontal tension enlarges angle <j> 
and the charges move in the opposite directions. Therefore the charge 




LA 



c,± 



*c 



•>->c k 



(a) 



(b) 





x ("ul 



(0 



Id) 



Figure 13.6 (a) Piezoelectric charge orientation; (b) crystal equivalent circuit; 
(c) crystal in the holder; (d) axes of the basic crystal cuts. 






320 



Oscillator Principles 



appearing on the upper and lower crystal surfaces alternately changes with 
variation of horizontal pressure. Conversely, if electrical charges are placed 
on the upper and lower surfaces by applying a voltage, a mechanical deforma- 
tion of the crystal occurs in the horizontal direction. 

An alternating voltage applied to the top and bottom crystal surfaces will 
cause the crystal to vibrate. If the time of travel of the mechanical vibration 
through the crystal is equal to a half cycle of the alternating voltage, then 
mechanical resonance occurs. The amplitude of the vibrations and the 
electrical voltage can be large. 

Rochelle salt is a very active piezoelectric material. It is easily damaged 
by moisture, however, and its use is confined to microphones. Quartz gives 
smaller output voltages but is mechanically very stable. The oscillation is 
electromechanical in nature and the frequency of resonance is dependent on 
a thickness dimension of the crystal. 

Electrical circuits employing piezoelectric crystals can be analyzed by 
replacing the crystal with its equivalent electrical network, Fig. 13.6(b). The 
magnitudes of L„ C„ and R, of the network depend on the way the crystal 
slice is cut and its thickness, as in Fig. 13.6(c). Capacitance C h is that of the 
mounting electrodes, usually electroplated onto the faces of the thin quartz 
slab. The valuable property of the quartz crystal is its sharp resonant response, 
which gives it a high equivalent Q. The Q is typically 30,000 but can reach 
500,000 in units mounted in evacuated cans to eliminate air damping of the 
vibrations. With a Q of 30,000, a crystal resonant at 4 MHz will have a 
bandwidth of only 133 Hz. 

The .v and y axes of the crystal are shown in Fig. 13.6(c), with x- and 
>'-cut crystals sliced perpendicular to the designated axes. These crystals have 
frequencies that vary with temperature but by an appropriate orientation of 
the angle of cut a zero temperature coefficient of frequency can be obtained. 
Resonant frequencies from about 10 kHz to above 20 MHz are possible with 
usable crystal thicknesses and higher frequencies can be generated as har- 
monic frequencies. 

Several crystal oscillator circuits are shown in Fig. 13.7. That in Fig. 
13.7(a) uses the crystal as a feedback clement, with maximum positive feed- 
back and oscillation at the crystal series or low-resistance resonance. The 
tuned circuit is adjusted near the crystal frequency and serves as an output 
waveform filter, the exact oscillation frequency being set by the crystal. 

Series resonance occurs when the left branch of the equivalent circuit 
appears resistive, at 

*-=&* <,3 ■ ,0, 

The resistance of this branch of the circuit at resonance is R, since the reac- 



.' 



Crystal Control ol Frequency 



Cc 



321 



Crystal 




V 



— WW 



tt 



ii 



Hf-o 



-/ 



-i Cr 



(a) 





Figure 13.7 (a) Transistor crystal oscillator, collector tuned; (b) FET-tuncd 
output oscillator. 

tances of L, and C. arc equal, opposite in sign, and cancel. This is the fre- 
quency of maximum positive feedback. 

The crystal electrodes form the capacitance C h with quartz as a dielectric. 
Capacitance C A causes a parallel resonance, with a resonant frequency f p 
determined when the reactances of the two branches are equal, as 

Xl. ~ Xc, = Xc, 

1 I 



2nf,L, - 
But this can also be written as 



2nf p C, 



(13.11) 



p«-i. 



2nf p L e m ^ ■!■ ^ 



2xf,\C* Cj 



f '_ 

" InjL.C 

where C is the equivalent series value of C, and C k , 



C. + C 



The effective circuit resistance is 



R - & 



(13.12) 



(13.13) 



(13.14) 



ajid is very high. The parallel-resonant frequency is very slightly higher than 



322 



Oscillator Principles 



The circuit in Fig. 13.7(b) employs the crystal in its parallel-resonant 
high-resistance mode as a resonant circuit in the gate, in a drain-tuned FET 
oscillator. Again, the tuned circuit supplies a frequency-selective filter but the 
crystal parallel resonance fixes the oscillating frequency. 

Changes in the load on the oscillator will alter the gain and thereby cause 
a change in the Miller-effect capacitance. This C„ is in parallel with the tuned 
circuit of an LC oscillator and shifts the frequency. The shift can be reduced 
by the use of large tank capacitance and small L in those circuits. Greater 
stability is given by use of a crystal since the Miller capacitance appears in 
parallel with C h . With an alteration of C h , which is in series with C,, the 
effective tuning capacitance C is not appreciably altered. In one example, a 
change of 10 per cent in C h changed the crystal frequency only 0.003 per cent. 

To isolate an oscillator from such changes in load that might affect the 
frequency, the oscillator is often followed by a buffer amplifier. This is 
designed to present a constant and high impedance load to the oscillator. An 
emitter follower is suitable for this purpose. 



13.5 Resistance-Capacitance Feedback 

Oscillator 

Variable LC oscillators tune over a frequency range that is proportional to 
l/VC. Available variable capacitors rarely have a maximum-to-minimum 
capacitance ratio greater than 10: I so that LC oscillators usually tune over 
ranges of VTO = 3.1. The function of the LC circuit is to provide the needed 
phase shift, however, and RC circuits can do this with a frequency range 
proportional to C, or over a 10: 1 frequency range. This decade frequency 
range is of advantage in laboratory oscillators, used in a circuit like that in 
Fig. 13.8. 



C»5! 




Figure 13.8 Basic Hewlett-Packard resistance-capacitance laboratory oscillator. 



Comments 



323 



An operational amplifier provides ~A gain internally, with 360° of phase 
shift ; the /? circuit of the series RC and parallel RC elements need provide no 
additional shift at the feedback frequency. With 



P- 



Z t /_0i +ZzL0i 



(13.15) 



z t + z t 

and the zero phase shift requirement can be met if 9, — 8 2 . With 

the zero phase shift occurs with /?, — R 2 — R and C, = C 2 = C at a 
frequency 



/o = 



1 



(13.16) 



2nRC 

The needed gain, with negative feedback provided by R 3 and R t in the 
configuration of the operational amplifier, is 



A -k^ 



(13.17) 



The negative feedback should provide a gain slightly greater than 3 so that 
oscillations will start. For instance, we might choose R 3 — 350,000 fl and 
R t = 1 00,000 fi. With negative feedback present to limit the gain close to 
the critical value, the oscillator does not operate far into cutoff and satura- 
tion; as a result, excellent sine waves can be obtained at the output. 

The R j, R A resistors are often changed to a scries resistor R 2 that supplies 
current to a low-power lamp as R t . Feedback is taken across the lamp and is 
small when the lamp is cold at low output. As the output increases, the lamp 
resistance increases much more rapidly than the current; the gain is reduced 
and the output regulated to a constant level. The gain is maintained just over 
the critical value of 3, cutoff and saturation are avoided, and excellent output 
waveform is obtained. 



13.6 Comments 

The oscillator serves as the source for high-frequency voltages. The feedback 
circuits discussed here are not the only forms but they are most generally 
used. 

The stability of oscillator frequency is the most important criterion of 
performance. The problem of stability of frequency would be a simple one if 
the oscillator were operated in isolation, at constant temperature, with 
unchanging components and constant voltages, and with no power taken 
from the circuit. Difficulties arise when we supply power from oscillators 
built with practical components and operating in normal environments. 



324 



Oscillator Principles 



At one time a crystal oscillator was almost the only means of obtaining 
stable operation; advanced materials and isolation of the oscillator function 
has now led to drift rates of only a few hertz per hour at megacycle fre- 
quencies, when using the tuned-circuit forms. 

REVIEW QUESTIONS 

13.1 Is the feedback negative or positive in an oscillator? 

13.2 What two requirements must be satisfied to make an oscillator from a feed- 
back amplifier? 

13.3 In a feedback amplifier, what is the minimum value of /? needed for oscilla- 
tion? 

13.4 What is the value of Aft for sustained oscillation? 

13.5 An amplifier has a gain of 5; how is this gain reduced to the limiting value 
when the amplifier oscillates? 

13.6 Explain the process of buildup of oscillations in an LC oscillator. 

13.7 Explain the process of buildup of oscillations in an Armstrong oscillator. 

13.8 What circuit determines the frequency of oscillation in a Hartley oscillator? 
The Colpitts oscillator? The Armstrong oscillator? 

13.9 How is feedback provided in an Armstrong oscillator? 

13.10 In what way is a crystal-controlled oscillator better than an LC oscillator? 

13.11 Why is quartz a suitable material for oscillating crystals? 

13.12 What is the effect of variation of output load on the oscillating frequency of 
an oscillator? 

13.13 How does output load affect the frequency of an oscillator? 

13.14 What is meant by the temperature coefficient of a quartz crystal? 

13.15 What is the phase shift requirement for the /? circuit of an RC oscillator? 

13.16 What gain must an amplifier have to be used in an RC oscillator? 

13.17 What is the purpose of using a tungsten incandescent bulb in the emitter or 
ground lead of an RC oscillator? 

PROBLEMS 

13.1 In Fig. 13.4(a), the Colpitts oscillator has C, = C 2 = C and L = 150 fiH. 
Find C, - C 2 for oscillation at 1.5 MHz. 

13.2 In a Colpitts oscillator in Fig. 13.4(a), C, = 250 pF, C 2 = 100 pF, and L 
- 300 ptH. What is the frequency of oscillation? 

13.3 A transistor Colpitts oscillator has L = 37 pH, C, = C 2 = 310 pF, Q = 
15, //„ = 5 x 10" 6 mho, and h lt = 800 Q. Find the frequency of oscillation; 
show the effect in hertz due to the correction term involving h ,lh„. 



Problems 



326 



/ 






13.4 
13.5 
13.6 
13.7 

13.8 
13.9 



In the circuit in Fig. 13.4(b), oscillating at 3.7 MHz, determine if C c = 0.05 
^F, L RfC = 1.5 mH, and C E =0.1 //F are suitable values. 

A quartz crystal has equivalent L, = 3.66 H, C, = 0.032 pF, C k = 6 pF, 
and R, = 4500 CI. What is the Q at the series resonant frequency? 

A quartz crystal has L, = 0.6 H, C. - 0.022 pF, C h = 5.42 pF, and Q = 
20,000. Find /„ and f„. 

The crystal of Problem 13.6 has a trimmer capacitor C m across C h to vary 
the oscillator frequency. If the trimmer is variable from 2.8 to 9.8 pF, find 
the possible range of variation of crystal frequency. (Use a calculator). 

A quartz crystal has L, - 250 H, C, = 0.04 pF, R = 1800Q, and C h = 7 

pF. When used to tunc the base circuit of a transistor with C lt — 30 pF, find 

the frequency of oscillation. The equivalent circuit is that shown in Fig. 

13.9(b). 

In Fig. 13.9(a) the circuit is to oscillate and C - 0.1 /iF, L = 0.15 H, and 

R, = 20,000 Si. 

(a) Find the gain A. 

(b) Is the phase requirement satisfied at resonance? 

(c) What is the value of /?? 

(d) What is the output frequency? 



50.000 n 
-WW 








::C h irC 



-°£ 



(b) 



Figure 13.9 



13.10 Choose the RC elements for an oscillator as in Fig. 13.8, for operation at /„ 

2 kHz. Minimum gain is to be 3.2. 

13.11 If R, = /? 2 = 10.000 Q, C t =Ci= 0.05 ftF, R 3 = 5000 Q, and R 4 = 
1500 £2, what is the frequency of oscillation of the RC oscillator of Fig. 13.8? 
What is the amplifier gain with feedback? 

13.12 In Problem 13.11, if M - 350, choose values for /? 3 and R t to assure that 
oscillations will just start. 

13.13 In Problem 13.12, C, and C 2 are changed to a split-stator or double variable 
capacitor with a range from 30 to 330 pF (each section). With R, ■ Ri — 
10 5 £2, what is the tuning range of the RC oscillator? 



14 



Modulation 
and Detection 



Electromagnetic radio waves propagate well through space only if the fre- 
quency is high, at least above 200 kHz. Speech and music frequencies lie in the 
band below 15 kHz; television picture signals utilize a video band ranging 
from 30 Hz to 4.5 MHz. To radiate such signals through space adequately 
requires that the base band frequencies be translated to appropriate channels 
in the radio-frequency spectrum that will carry the signals. These channels 
presently cover the frequency range from 200 kHz to many gigahertz. Such 
translation or conversion of frequency is accomplished by the process of 
modulation. 

Detection, or demodulation, is the name for the reverse process by which 
the desired signals are recovered from the radio-frequency carriers and made 
audible or visible as in television. 



14. 1 Fundamentals of Modulation 

An alternating voltage can be expressed as a function of time as 

v = A cos {In ft + B) (14.1) 

where 

A ns peak amplitude of the wave 
/= frequency in hertz 
6 — phase angle 
t = time 



326 



The Frequency Spectrum In AM 



327 



In telephone and radio transmission we want to convey information to 
ihe receiver. The wave of Eq. 14.1 can carry information only through its 
presence or absence. The telegraph is an example of such an on-off signal but 
the telegraph has been limited in its rate of transmission of information. To 
send information as fast as it is generated, or in real time, requires that a 
characteristic of the radio wave be varied at the real time rate. 

Equation 14.1 has two characteristics capable of being varied with time in 
accordance with the information we want to transmit. Thus we have two 
basic methods of modulation of an ac wave, as 

1. Amplitude modulation (AM), in which the wave amplitude A is caused 
to vary in accordance with the amplitude of the modulating signal. 

2. Frequency modulation (FM), in which the frequency / of the wave is 
changed in accordance with the amplitude of the modulating signal. 

Actually, FM is a subproccss of a more general form known as angle 
modulation, as is phase modulation (PM) in which 6 is caused to vary with the 
modulating signal; however, FM is most generally used. 

Another class of systems does employ the method of the telegraph but 
turns the signal on and off at a very high rate and generates very short pulses 
for transmission. The information signal is sampled at a rate of several 
thousand per second and a characteristic of the pulse is varied to represent 
the amplitude of each sample. We have several possibilities: 

1. PAM, pulse-amplitude modulation, in which the amplitude of the 
pulse is varied by the sample amplitude. 

2. PDM, pulse-duration modulation, in which the duration of the pulse 
represents the sample amplitude. 

3. PCM, pulse-code modulation, in which a coded train of pulses repre- 
sents the sample amplitude. 

The PCM system is becoming widely used because of its freedom from 
noise and distortion in the transmitting path, wire, cable, or space. In the 
reception of PCM we do not need to receive an accurate pulse waveform. By 
transmitting the sample amplitude in a code of pulses, we need only to deter- 
mine that a pulse was sent or not sent. The binary code of Chapter 16 is 
generally employed in generating the pulse trains, which are decoded back to 
sample amplitudes at the receiver. 



14.2 The Frequency Spectrum in AM 

The frequency of the radio-frequency wave will be designated f c , as the car- 
rier frequency on which the informational signal is to be modulated. The 
informational signal from speech, music, or the dots of a TV picture will be 



328 



Modulation and Detection 



assigned a frequency/,. The frequency/, will be only one of a large band of 
frequencies making up the complete music or TV picture spectrum; we use 
this one frequency as an example of what happens to every such frequency. 
The signal frequency/, is smaller than the carrier frequency/ on which it is to 
be modulated. 

The signal frequency can be written as 

v, -- V, cos 2b/,/ (14.2) 

and the carrier frequency on which we wish to modulate the information 
signal is 

v c - A cos 2b// (14.3) 

from Eq. 14.1, after dropping the constant angle as having no meaning here. 
In amplitude modulation we vary the coefficient A with the informational 
signal so that 

A = V c |- V, cos 2b/,/ 
= K,(i+£cos2b/,/) <"•<> 

The modulation factor is 

«.=£ (14.5) 

and 

A = V c (\ -\- m a cos 2jif,t) (14.6) 

In AM systems it is not desirable for m„ to exceed 1 .0 or 100 per cent because 
of excessive distortion that is generated. 

Substitution of Eq. 14.6 for A into Eq. 14.3 gives us an expression for the 
amplitude-modulated wave: 

v = V c {\ + m a cos 2b/,/) cos 2b// (14.7) 

= V c cos 2b// I m„V c cos 2b// cos 2b// (14.8) 

The first term is the carrier and the second term can be simplified if we use 
the trigonometric identity 

cos a cos b = \ cos (a -f- b) + £ cos (a — b) 

showing that the second term actually consists of two waves. We can then 
state the amplitude-modulated wave as 



v = V e cos 2b// + '-f V c cos 2b(/ + /)/ + y" V c cos 2b(/ - /)/ (14.9) 

The amplitude-modulated wave consists of three frequencies, the original 
carrier at/ and two side frequencies. The upper side frequency appears as the 
sum of the carrier and modulation frequencies,/ +/, and the lower side 
frequency appears as the difference of the carrier and the modulation fre- 



The Frequency Spectrum in AM 



329 



quencies,/ - /. If/ is small, then the three frequencies are closely grouped 
and centered on/. 

The resultant waveform is shown in Fig. 14. 1 , with the constant amplitude 
carrier in Fig. 14. 1(a) and the modulated wave in Fig. 14.1(b). The modulated 
wave is not the result of simple addition of two frequencies but is the sum of 
three frequencies of Eq. 14.9. Amplitude modulation occurs because of the 
product of two frequencies as found in Eq. 14.8. Such a product is aneccssary 
condition for any method of amplitude modulation. 

If the modulating signal is one of many frequency components in speech 
or music, as examples, then many side-frequency pairs exist, and the groups 
of side frequencies are called the upper and lower sidebands. Figure 14.2(a) 
shows the spectrum of an AM waveform in which a carrier at / has been 
modulated by a frequency/, generating two side frequencies. In Fig. 14.2(b) 
we have an AM spectrum in which a carrier at/ is simultaneously modulated 
by three signals at 1000, 2000, and 4000 Hz. 

The information-carrying signal has been translated by amplitude modula- 





ta) 

Figure 14.1 (a) Unmodulated waveform; (b) ampliiude-modulatcd wave, 
m fl = 0.5. 



Carrier 



Carrier 



Lower Sideband 



jj. 



Upper Sideband 



J_L_L 



fc-f, fc fc+f, 
(a) 



-4k lk/ c +Ik +4k 

2k +2k 

(b) 

Figure 14.2 Frequency spectra for AM modulation. 



330 



Modulation and Detection 



tion to a different frequency / c , ideally without distortion. The carrier/, can 
be placed anywhere in the radio spectrum for ease in transmission over wires 
or through space or for better conditions of amplification. The bandwidth 
occupied by the signal has been doubled since we now have two sidebands. 
That is, the modulated wave represented by the spectrum in Fig. 14.2(b) with 
f t = 1 MHz would have sidebands extending from 0.996 to 1.004 MHz, a 
total bandwidth of 8 kHz and twice the frequency of the highest-frequency 
modulating signal. With frequencies scarce in our crowded radio bands, this 
doubling of bandwidth is one of the deficiencies of the AM system of modu- 
lation. 

The waveform of an AM wave would appear on a cathode-ray oscillo- 
scope as in Fig. 14.1(b). Measurements A and B may be made to find the 
value of m„ from the pattern; that is 






m a =» 



A- B 

A B 



(14.10) 



14.3 The Power Spectrum in AM 

Equation 14.9 tells us that the carrier has an amplitude V r and when the 
modulated wave is applied to a resistive load R, the power due to the three 
components in double sideband AM (DSB) is 
Carrier power: 

(14.11) 



(14.12) 





te R 


Upper sideband: 






p ml V\ ml p 

' 4 R ' ' 4 c 


Lower sideband: 






p _ ml V} _ ml p 

4 R -- 4 c 



(14.13) 



(14.14) 



The total power in an AM wave is therefore 

P-P.(\ I f) 

The total power in the sidebands is ml/2 limes the power in the carrier and is 
divided among the many frequency components, each frequency having its 
own value of m. such that 



m: 



ml. 



m- 



mi 



At m„ = 1 .0 or 1 00 per cent modulation, the sideband average power with 
sinusoidal modulation is 50 per cent of the carrier power and the total average 



The Diode Modulator lor AM 



331 



power is 150 percent of the unmodulated carrier power. The signal sidebands 
employ only one-third of the total power of an AM wave at lOO per cent 
modulation, the other two-thirds being present in the carrier. 

In Fig. 14.1(b) the peak above V c \%mV c and at m - 1.0, the peak voltage 
is 2V S and double the carrier level. The equipment must be designed to with- 
stand such a voltage at peak modulation. 

With 2V C at the peak on full modulation, the power is 4V-/R = 4P C and 
power peaks of four times the normal unmodulated carrier must be supplied. 

A radio station is rated at a carrier power of 1000 W. To reach 100 per 
cent modulation we must supply 500 W of average modulating power and 
there will then be 250 W average in each sideband. At 10 per cent modulation 
i ma -_ 0.10) the carrier power is still 1000 W but by Eq. 14.14 the average 
sideband power is only 2.5 W. Since only the sidebands are usable power at 
the receiver, this low modulation percentage gives a very poor power effi- 
ciency for information transmittal. Accordingly, the modulation factor m„ is 
normally maintained near 1.0. 

14.4 The Diode Modulator for AM 

Amplitude modulation is often generated by use of the nonlinear voltage- 
current curve of a diode or that of the transistor emitter-base junction. Let 
us approximate the diode curve in Fig. 14.3(a) by use of a linear curve added 
to a parabolic curve, as 

/ = fl ,« -\ dzV 1 (14.15) 

In Fig. 14.3(b) two voltages arc applied to the diode and load circuit, resonant 
at frequency f c : 

v=V c cos 2nf c t -f V, cos 2nf,t (14.16) 




1 


M 








1) 






ft© 




Tu 




/■ — s. 




C 




o ,u 


o 
o 




o 




x ti 




o 


'■(: 







(a) 
Figure 14.3 



(b) 



(a) A diode characteristic; (b) simplified diode modulator. 



■ 



332 



Modulation and Detection 



where V, < V c and f, </ c . Substituting this voltage expression into Eq. 
14.15, we can develop an equation for the circuit current as 



(14.17) 



i = a, V c cos 2nf c t + a, V, cos 2nf,t -|- a,V\ cos 2 2nf c t 
+ a 1 V) cos 2 2nfj - 2a 2 K c K, cos 2jr/ J r cos 2n/ c / 
By trigonometric identity, we have 

cos 2 a = \ cos 2a + £ 
and the third and fourth terms of Eq. 14.17 can be modified to 

a % V\ cos 2 2nf c t m ^p[cos 2n(2f c )t + 1] 
a 2 KJ cos 1 2nf,t - ^[cos 27r(2/> + i-] 

But 2/ and 2/, are double frequencies or second harmonics. 

Rearranging Eq. 14.17 and using these equations involving the second- 
harmonic frequencies, we have 

i - a, V c cos 2nf c t + ^p[oos 2ti(2/ c )/ + JU 

+ a, K, cos 2b/,/ + ^[cos 2n(2f,)t + ^-1 
+ 2a 2 V e V, cos 2r/,/ cos 2»/ c / 



(14.18) 



This current passes through the tuned circuit, resonant and having a high 
resistance only near frequency /. The frequency 2f c is far removed from/, 
being a second harmonic, and so are frequencies/, and 2/ as modulating 
frequencies, and the dc terms a 2 K c 2 /2 and a 2 K;/2. Only the first and last 
terms of Eq. 14.18, involving frequency/, will produce appreciable voltages 
across the resonant circuit. Thus that circuit acts as a filter to remove the fre- 
quencies we do not want in the modulator output. 
Therefore, the effective output voltage is 

v = a,/f,K c cos 2nf c t -I 2a 2 R„V,V c cos 2nf,t cos 2nf c t (14.19) 

where R p is the resonant impedance of the circuit. This expression shows the 
frequency product term predicted as necessary for amplitude modulation. We 
can reduce Eq. 14.19 to 

v = a , R P V C ( 1 -• -^ V, cos 2nf,t) cos 2nf c t (14.20) 

which shows the voltage in the form of Eq. 14.7 as an amplitude-modulated 
wave. 

This process is known as a small-signal or low-power method of modula- 
tion. 



High-Power-Level AM Modulation 



333 



14.5 High-Power-Level AM Modulation 

A method of amplitude modulation better suited to high-power use is that of 
power conversion in the Class C modulated amplifier shown in Fig. 14.4. With 
the input at/, the high-frequency current output is linearly related to the 
supply voltage to the amplifier. If that supply voltage is varied by a Class B 



V c cos 2itf c t 



Modulated 
Amplifier 





Class B 
Modulator 






Signal 


o 
o 


at/, 




o 

■ 












Modulated 
Output «r 



'4 



/c 






V t cos 2jt/,/ 



V cc 



fa) 



Output 
Current 




Figure 14.4 (a) High-level AM modulator circuit ; (b) resultant output current. 



334 



Modulation and Detection 









amplifier as modulator, wc have 



». = Vcc + V, cos 2nfj 
= V cc (\ + m.cos2nf,i) 



(14.21) 



The current lo the output tuned circuit varies in a similar manner, giving 



/ = -^(1 I m, cos 271/,/) cos 2nf c t 



(14.22) 



and this represents an amplitude-modulated wave, with carrier at f c and 
sidebands/. • /, and f c — f„ Eq. 14.7 and 14.9. With the supply voltage 
varying at the/, rate, then the carrier frequency current in the lank circuit 
will have an envelope shape corresponding to the modulating signal. 

The current from the dc supply is I c since the modulation-frequency 
variation is up and down from the steady level and averages out. The dc sup- 
ply furnishes the steady carrier power K cc ./ C and the modulator must furnish 
the power for the sidebands, at maximum being V cc l c /2. The output trans- 
former of the Class B amplifier must be adequate for this level of power. 

The secondary load resistance for the Class B modulator must be known 
in order lo specify the turns ratio of the transformer. The resistance into 
which this transformer delivers the sideband power is that of the modulated 
amplifier at the modulation frequency. This is 



„ _ modulation component of secondary voltage 
* modulation component of secondary current 

^ m a V cc cos 2nf,t _ V cc 
m a I c cos 2nf,i I c 



(14.23) 



which is simply the dc resistance represented by the modulated amplifier. 
Another form of amplitude modulation forces a current of one frequency 
through a resistance or impedance whose magnitude is varied at a second 
frequency. Thus 

v = /, cos 2nf,i x R cos 2nf c l (14.24) 

This shows directly the frequency product term needed for amplitude modula- 
tion to take place. This method is used in low-frequency servo systems by 
variation of an inductance at the carrier rate. 



14.6 Linear Detection for AM Signals 

To recover the useful information from an AM wave we normally use a 
diode detector or envelope demodulator. The circuit in Fig. 14.5(a) acts much 
as a half-wave diode rectifier with a shunt-capacitor filter. 

Capacitor C is chosen so that the parameter//?C is in the range of 30 to 



Linear Detection for AM Signals 



33S 



P 



-* — r 



u C'ss R 




— i 



AVC 

(a) fb) 

FiKure 14.5 (a) Linear diode detector circuit; (b) output waveform. 

200, as done for the rectifier. The capacitor charges to the peak of each carrier 
cycle and holds that voltage until the next positive half cycle. With an ampli- 
tude-modulated wave as an input signal, 

v = V&\ + m. cos 2nf,t) cos 2nf c t 

The voltage of the capacitor at the carrier peaks is 

V. = V c {\ + m a cos 27i/./) (14.25) 

as shown in Fig. 14.5(b). Since m„ = V,\V e> we have 

^ = r«+^cos27r/,/ (14.26) 

This shows that in the voltage across the load R wc have recovered the origi- 
nal modulation signal and also have a dc term equal to the carrier amplitude 
of the received signal. 

The voltage K„ appears across R and C c blocks the dc component but 
passes the modulation frequency /, to the output of the detector. Practical 
circuit values are shown in Fig. 14.6 for/, * 2 MHz. 

While R and C are chosen so that/ c /fC is large, they should be chosen so 
that/,/?C is approximately given by 

I 



max/flC < 



2nm a 



(14.27) 



Such a choice ensures that RC is small enough that at frequency/ the capac- 
itor C can discharge between cycles of/, and thus its voltage is able to follow 




Figure 14.6 A practical detector circuit for AM. 



336 



Modulation and Detection 



changes in V„ the information signal amplitude. Distortion results if Eq. 
14.27 is not approximately satisfied. 



14.7 Automatic Volume Control (AVC) 

In radio reception it is desired that weak signals be amplified more and 
strong signals be amplified less so that the output level of all signals is about 
the same and remains so even though signals/orfe and weaken in transmission. 
The dc term in Eq. 14.26, equal to the received carrier level, provides a 



Signal 



Output 




Circuit — 



Output 



(a) 



V 






r/ 






-°/ 


Detay^ 


AVC 




"""ob* 6 


Zv<f 


* i 







Delay 
Level 



Signal y c 



(b) 



Figure 14.7 (a) An AVC circuit; (b) automatic volume control action. 



The Single-Sideband System of Modulation 

measure of received signal strength that is used to achieve automatic volume 

'"shovvn^lhe circuit in Fig. 14.7(a) is an * a C, filter at the diode detector 
nutoul The R Z C 2 product is made large, a fraction of a second, to remove 
°ny modulation at low/, frequencies from the dc voltage. Connected as 
'shown, the output or D t makes this dc voltage negattve to ground After 
filtering, this dc voltage is applied to the base of the npn transistor Q t and 
other amplifier transistors, to reduce l E . Both h„ and h„ are functions of l s 
and the gain of the amplifiers can be reduced as the strength of the received 
signal increases, as measured by V c . 

' Without AVC the output of the detector increases in proportion to the 
input signal as shown in Fig. 14.7(b). The AVC action decreases the gain with 
increased signal strength, giving an output change as shown by the dashed 
curve and holding all signals more nearly constant in output. 

All signal levels produce an AVC voltage, however, and the gain is reduced 
even for weak signals, where the full gain is needed. Accordingly we use 
delayed A VC, applied by the diode clamp circuit at point A on the AVC line 
of the amplifier. Until the AVC line reaches - V„, the delay toas diode D 2 is 
closed and transmits -V B volts on the AVC line; this is the des.red bias 
level for all amplifiers. But when the AVC voltage from D, becomes more 
negative than -V„, diode D t opens and the varying AVC bias is used to 
control the amplifier gain. The result is a high gain for weak signals and a 
more ideal AVC characteristic for large signals, as shown in Fig 14.7. 

14.8 The Single-Sideband System 

of Modulation 

At 100 per cent modulation in an AM signal, the carrier requires two-thirds of 
the power but conveys no useful information. The carrier can be viewed as a 
power waste and the second sideband as needless duplication, except for its 
rare aid when selective fading distorts one sideband in radio transmission. If 
we use only one sideband and suppress the carrier, we have the system gener- 
ally known as single-sideband (SSB) transmission. We are able to transmit the 
information with reduced power requirements, half of the bandwidth and 
lessened interference between signals, but at some expense in complexity ot 

equipment. . 

A carrier must be introduced at the receiver, closely adjusted to the origi- 
nal carrier frequency since a A/ carrier difTcrence produces a A/ shift in all 
signal frequencies and introduces distortion. The carrier must be accurate to 
within 10 or 20 Hz for intelligibility of voice signals, and stable oscillators are 
required to generate the local carrier. 

The first step in generation of an SSB signal is to develop an AM signal 



338 



Modulation and Detection 






t" 



ih- 



000 






v2) 



'Output 




(a) (b) 

flfeHK /<.* (a) A balanced modulator; (b) a diode balanced modulator. 

without carrier. A balanced modulator is used, as shown in Fig. 14.8(a). The 
respective inputs to Q, and Q z have the/, signal common and the/, signal 
differentially connected, giving 

v, = V c cos 2nf c t + V, cos 2nf,t (14.28) 

f j = K c cos 2nf e t - V, cos 2nf,t (14.29) 

The emitter-base junctions of the transistors can be assumed to have volt- 
age-current relations that can be represented by Eq. 14.15, as 

/ = a,v f a 2 v z 

Use of Eq. 14.28 and 14.29 as the applied voltages produces mixing of the two 
frequencies; with the push-pull connection giving a subtractive output, the 
output expression will contain only the terms 

/. - 2a , V, cos 2nf,t -\ *a z V,V c cos 2nf,i cos 2tt/ c / (14.30) 

Now in Sec. 14.2 we used the trigonometric identity 

cos a cos b — $ cos (a -|- b) I J cos (a — b) 

Expanding the product term of Eq. 14.30 according to this relation, we have 

/„ = 2a, V, cos Infjt + 2a 2 K,K c [cos 2n(f e +f,)t | cos 2n(f c - f,)t] 

The output circuit is parallel resonant and of high resistance R p only near 
frequency/,. Frequency/ </ c by assumption and the first term does not 
produce an appreciable voltage across the circuit because the circuit imped- 
ance is negligible at/,. Thus, with R p as the resonant impedance, the vojtage 
across the output circuit is only 

v = 2o 2 fl ; ,K 1 |/ c [cos 2n(f c +f,)t ■ cos 2n(f c - f,)t) (14.31) 
where f t +/, and / - / represent the sidebands of an AM wave and with 






Frequency Translation; the Product Detector 339 

the carrier absent. This result should not be surprising since the carrier rep- 
resented a common input to the two transistors, in push-pull or differential 
connection. 

Another form of balanced modulator is shown in Fig. 14.8(b), using 
diodes instead of transistors. The action of the circuit is similar to that of the 
transistor circuit and the output consists solely of the two sideband frequen- 
cies. The circuit is commonly used in telephone carrier-current transmission 
and requires accurate center-tap connections on the transformers for balanc- 
ing out the carrier. 

The sideband outputs are shown schematically in Fig. 14.9(a). A sharp 
cutoff filter must be used to select one of the sidebands, yielding the SSB 
signal in Fig. 14.9(b). The filter must separate signals in the two sidebands 
that differ only by twice the lowest modulation frequency. For voice frequen- 
cies the amplifiers arc usually designed to cut off at about 200 Hz so that the 
lowest frequencies will differ by at least 400 Hz, as indicated by the separation 
of the bands in Fig. 14.9(a). Filters that can achieve this amount of skirt 
selectivity and place the rejected sideband at least 30 dB below the level of 
the accepted frequencies are usually designed with piezoelectric quartz ele- 
ments of high Q and operate at/ c frequencies of 2 to 5 MHz. 





(a) 



(b) 



Figure 14.9 (a) Balanced modulator output spectra; (b) output after lower 
sideband is removed by filtering. 



14.9 Frequency Translation ; 
the Product Detector 

By use of the principle of modulation it is possible to translate a band of 
frequencies centering at f„ to a band of similar frequencies centering at an- 
other frequency f b . As for modulation, an input signal of frequency f e and 
a locally generated signal at /, are simultaneously applied to a diode, tran- 
sistor, or tube having a parabolic characteristic to produce the frequency- 
product term necessary for modulation. An output voltage then appears in 
a circuit tuned to/ e ±f x - 






340 



Modulation and Detection 



The signal to be translated may be an AM wave, having a carrier/ and 
side frequencies at A -\- f, and/ — /,. The translation process yields sums 
and differences of all input frequencies and if we mix a local signal at fre- 
quency/,, we shall have the following output frequencies: 

/. A 

f. I- /.-/, A-A=A 

f. I A + A =/, + A A +/. ~f„ - A +/, 

/. -/. i- A -A -A A -A -A -A -A 

The sum and difference terms will have amplitudes proportional to the 
products of the individual wave amplitudes. 

The frequency groups centered at A and A constitute two translated AM 
waves, each containing the original side frequencies but moved up and down 
onto new carrier frequencies. This principle of frequency translation receives 
wide application in receivers and transmitters. 

Additional frequencies of small amplitude are generated in the process. 
These spurious frequencies can create interference signals in other channels 
unless suppressed by high-g resonant filters. 

The translator in Fig. 14.10 employs one transistor for local generation of 
A and the mixing of the frequencies. The resonant circuit of L x , C x and the 
feedback coil L M form an emitter-tuned oscillator operating at f x . The 
internal interaction through the base-emitter junction with f. produces an 
output in the L k , C k circuit tuned to/. The A output is predicted by a con- 
version transconcluctance g c and the conversion gain is 

A c - g c R (14.32) 

where R is the resonant resistance of the output circuit at frequency/. The 
circuit is useful as a translator even if A c is less than unity. 

The process of frequency translation is also employed in the product 
detector by which we reinsert the carrier and recover the SSB signals. Suppose 
that the input at/ in Fig. 14.10(b) is modulated as an upper sideband/ + /• 
Then suppose that the locally generated oscillation is adjusted to f x /].. 
The mixed output will include frequencies at 

A 

A+A 
A +A +/. 
A -t-A -A -/. 

The first three output frequencies are high and are bypassed by C but the 
last frequency at A ' s tna t of the original modulation; it is passed to the 
output through C c . 

This is the method for reintroduction of the carrier, and for good intelli- 
gibility the local oscillation must be within 10 or 20 Hz of the/ used in gen- 
erating the transmitted sideband. Severe requirements are seen to be placed 
on the stability of the local oscillator. 



The Frequency Spectrum of FM Signals 



341 







(a) 




■f [ Audio 
Outpul 



Figure 14.10 (a) Transistor frequency translator; (b) product detector. 



14. 10 The Frequency Spectrum of 

FM Signals 

Most natural and man-made radio noise is in the form of amplitude variations 
of the signal and a system that eliminates amplitude variation in its received 
signals can also eliminate most radio noise. The system of frequency modula- 
tion operates with constant amplitude signals and is effective in reducing 
noise in radio reception. 

If we are to design circuits for FM equipment, we must know the fre- 
quency spectrum required by the frequency components of a frequency- 
modulated (FM) wave. We shall again use as the information to be 
transmitted a frequency/ and a signal voltage 

v, = V, cos Inf.t (14.33) 

We vary the frequency of our FM wave in proportion to the amplitude of 
this signal. 

We define the frequency deviation as 

ft**kfV t (kHz) (14.34) 




342 



Modulation and Detection 



where k, is the proportionality factor in kilohertz per volt. This relates the 
amplitude of the modulating signal to the frequency variation of the FM 
wave. 

Suppose that a 1-V signal produces a IO-kHz deviation, then k f = 
10 kHz/V. This is applied to a 50-MHz frequency and the frequency is shifted 
to 50.01 MHz. A 2-V signal produces a 20-kHz deviation to 50.02 MHz; a 
— 2-V signal swings the frequency to 49.98 MHz. If the frequency / is 
1000 Hz, then the swings between 50.02 and 49.98 MHz occur 1000 times 
per second for a 2-V peak cosine signal. 

Maximum values of/, are assigned for various radio services, as 75 kHz 
for sound broadcasting and 25 kHz for the sound channel in television. 
Maximum signal frequencies are/, (ra ,„ = 15 kHz for the audio range. Then 
we define the deviation ratio as 



Jilniil /i(mtx) 



(14.35) 



Thus m, ■= 75 kHz/15 kHz — 5 for sound broadcasting. 

With modulation by the signal of Eq. 14.33, our FM wave can be written 
as 

v = V a cos (2a/,/ -i- m f sin 2a/,/) (14.36) 

The frequency/, is called the center frequency because the FM signal deviates 
up and down from the/ value, in accordance with the modulation term in 
the parentheses. The center frequency may go to zero for some values of m s 
and this is the reason we do not call / a carrier frequency. There are side 
frequencies in pairs above and below the center frequency at every harmonic 
of/ all the way to infinite frequency. The side frequencies are not restricted 
to ±Z as for AM. 

Practically, we do not have to contend with FM signals of infinite band- 
width. The amplitudes of the high-order side frequencies decrease quite 
rapidly and become negligible; however, we do use bandwidths of ±75 kHz 
for FM sound broadcasting. The wider bandwidths give greater suppression 
of noise in FM transmission through space. 



14. 1 1 Bandwidth of FM Signals 

The necessary FM bandwidth for good waveform reproduction can be 
assumed as 

BW ~ 2/ (n ,„, (14.37) 

Spectrum frequencies and amplitudes for an unmodulated amplitude of 
unity are plotted in Fig. 14.11 for a variety of signal and m, conditions; all 
components with amplitudes over 1 per cent are shown. 



0.76 






./> 



./> 



L 



J 



•fd-* 



-Id- 



\L. 



^1 



(a) 

f d = 5 kHz 
/, =5 kHz 
in, = 1.0 



(b) 

l a = 25 kHz 

/, =5 kiiz 

m, - S.O 



f d =50 kHz 
/, = 5 kHz 
m, = 1 0.0 



M) 



r« 


= 75 kHz 


A 


■ 5 kHz 


'"/ 


= 15.0 



Lu 



— 75 kHz — -\- 75 kHz— 

/„ 
Figure 14.11 Spectra of FM waves for various m f values. 



343 



344 



Modulation and Detection 



In FM broadcasting, f i{m , x) has been established at 175 kHz and the 
usual design makes the receiver bandwidth ± 100 kHz, which accommodates 
all side frequencies that arc greater than 1 per cent in amplitude, according to 
Fig. 14.11(d). 

Since the amplitude V B is unchanged by frequency modulation, the average 
powers in unmodulated and modulated waves are equal. Modulation simply 
spreads the available energy among the sidebands. 

The spectrum for m f — 1.0, shown in Fig. 14.1 1(a), closely approaches 
that of an AM wave and it is found that for m f = 0.6 or less the spectrum 
reduces to 

v„ = V B [A„ cos 2nfj + A , cos 2n(f„ + /,)/ - A , cos 2n(f - f,)t] (14.38) 

which is similar to double-sideband AM. The bandwidth is then 

bw = y JtaMj 

and the system is said to be one of narrow-band FM. For m f > 0.6, the 
system is considered wide-band in nature. 



14. 12 Generation of FM Signals 

An FM signal may be generated by causing the capacitance of a tuned-circuit 
oscillator to vary with the modulating signal amplitude. A semiconductor 
diode may be connected across the tuned circuit, with the diode under reversed 
bias. The capacitance of the junction will vary if a bias voltage 

»* — - V BO I V, cos 2nf,t (14.39) 

is applied. With the capacitance of the diode proportional to the squaie root 
of the applied voltage, the amplitude of V, must be kept small for linearity. 
But with V, small, the frequency of the oscillator will be 



and this conforms to the needs for FM generation. 



(14.40) 



14. 13 The Limiter and Discriminator 
for FM Detection 

Noise and varying amplitude interference may arrive with the FM signals. 
Since FM signals have constant amplitude, this interference may be removed 
by limiting all signals to a common amplitude. With this done, the only vary- 
ing property of the input is frequency and we can proceed to convert the 



V 



The Limiter and Discriminator for FM Detection 



345 



frequency variations back to the amplitude variations of the original speech 

or music. 

One form of amplitude limiter employs an amplifier with low collector 
voltage so that the load line of the transistor is short and all incoming signals 
arc large enough to drive the amplifier to cutoff and to saturation. Since this 
distance on the load line is fixed, all signals above a threshold value will ap- 
pear at a uniform output level; this is illustrated in Fig. 14.12(a). 



-<s 



-»: 




-Useful Range- 



Noisc Output 



10 20 30 

Limiter Input (V) 



40 





0.8 




0.6 




0.4 


s? 


0.2 


1 





u' 


0.2 




-0.4 




0.6 




0.8 



(a) 



200 - 100 100 200 

k'.:/ Deviation at 10 MHz 

(b) 

Figure 14.12 (a) Limiter performance; (b) discriminator action. 



Another circuit that adopts operational amplifier construction is shown 
in Fig. 14.13(a). The transistors arc normally biased into equal conduction. 
For any input V, more negative than some value -V x , g, is cut off and the 
bias from R E is reduced because of the smaller common current. This reduc- 
tion in bias raises the current in Q 2 and its output voltage becomes V, u . 
Then as V, rises, Q x begins to conduct and the emitter voltage rises, reducing 
the current in Q x . At some positive V, = +V X on Q„ the bias across R B 
becomes large enough to cut off Q x and its output voltage goes to V cc . 

For any input signal with peak-to-peak amplitude exceeding 2V„, the 
output is limited at a peak-to-peak value of V^ - V m . This action is demon- 
strated in Fig. 14.13(b). 

The Foster-Seeley discriminator shown in Fig. 14.14 is one form of detec- 
tor for FM signals. The input transformer serves as a load for a limiter circuit 
and the primary voltage is in scries with the secondary voltage to ground, the 
reactance of the blocking capacitor C being neglected. The diodes D„ and D b 
have applied voltages 

(14.41) 



V„ = V x + % 



-V -£ 



The primary and secondary voltages of the overcoupled circuit, at a 



346 



Modulation and Detection 




v G, ore 




(a) 



(b) 
Figure 14.13 An emitter-coupled limiler. 

2 




Limiter 



Discriniinutor 
Figure 14.14 A frequency discriminator. 



resonant frequency equal to the FM center frequency, are 



(14.42) 
(14.43) 



The./ factor in V z shows it to be in phase quadrature to V x . Thus we draw the 
phasor diagram in Fig. 14.15(a) with V, and K 2 at right angles at resonance. 

Diode voltages V a and V„ are shown as the sum and difference of K, and 
V x \2 according to Eq. 14.41. At frequencies below resonance the phase of K 2 
changes toward the diagram in Fig. 14.15(b) and above resonance the phase 
of V t shifts oppositely toward the diagram in Fig. 14.15(c). 

The diodes provide an output equal to V„ — V b and at resonance this is 



The Ratio Detector for FM 



347 



2 





(a) 



(b) (c) 

Figure 14.15 Phasor description of discriminator performance. 



zero. At deviations above and below resonance the angle 9 becomes smaller 
and larger, respectively, and the difference V a — V b progressively changes. 
This voltage difference is translated into the diode output curve in Fig. 
14.12(b). Linear output voltage versus frequency is available for a range of 
nearly ±150 kHz at a center frequency of 10 MHz. 

Equation 14.42 shows that the output is proportional to the input ampli- 
tude K„; thus the output would vary with amplitude or noise if the circuit 
were not preceded by a limiter. 

The transformer is overcoupled and the separation of the response peaks 
in Fig. 14.12(b) is dependent on the Q of the primary and secondary circuits. 
With Q t = Q 2 — Q, the needed value is 



= 



L 



2/,J(m.i) 

For speech and music 2f dla , x) = 200 kHz and at/ = 10 MHz, the value of 
Q will be 50. 



14. 14 The Ratio Detector for FM 

Another detector for FM is the ratio detector drawn in Fig. 14.16. The full 
secondary voltage charges C c to the peak of signal V, through the two diodes 
in series. The values of R 2 and C c are large so that V is constant for a given 
signal but varies for different signal amplitudes. 

As in the discriminator, the primary voltage and one-half of the secondary 
voltage is applied to each diode. At the center frequency these voltages are 
equal and there is equal charge in C, and C 2 but there is no voltage across 
/?,. This is true regardless of the magnitudes of V a and V b . When frequency 
deviations occur, the unequal diode voltages charge C, and C 2 unequally but 
the total of V„ and V b is constant at V. Voltage at A is dependent on the 
proportional change in V a and V b and so we have the ratio detector. A varying 
signal then appears at A to ground. 



348 



Modulation and Detection 




Figure 14.16 The ralio detector. 

The performance of the circuit is somewhat affected by signal amplitude 
and the operation is improved if the circuit is preceded by a limitcr. 



14. 15 Automatic Frequency 
Control (AFC) 

At the high frequencies employed with FM a crystal cannot always be used 
and automatic frequency controI(AFC) circuits are used to improve frequency 
stability. 

A discriminator circuit is connected to the oscillator output as in Fig. 
14.17. When the oscillator frequency corresponds to the center frequency for 
which the discriminator is tuned, the output from the discriminator is zero. 
When the oscillator is off frequency, a positive or negative voltage is devel- 
oped at the discriminator output. This voltage is filtered to remove signal 
components and applied to a reverse-biased diode capacitor across the oscil- 
lator tuned circuit, where the voltage adds or subtracts to a fixed reverse bias. 
A bias change in one direction causes an increase in diode capacitance and a 
reduction of oscillator frequency; a bias change in the opposite direction 
causes a reduction in diode capacitance and an increase in oscillator 
frequency. 



Biased 




Oscillator 




Amplifier 


» Output al/„ 


Diode 


























Filter 





















Figure 14.17 Automatic frequency control (AFC). 



Review Questions 



349 



The directional bias from the discriminator is always so polarized as to 
shift the oscillator frequency toward its proper frequency. FM receivers are 
stabilized against frequency drift with temperature by such circuits. 



14.16 Comments 

The transmission of information by radio waves requires that those waves be 
controlled by the informational signal. We can vary any one of three charac- 
teristics of the radio wave, its amplitude in AM, its frequency in FM, and its 
phase in PM. Thus low-frequency audio and picture signals can be transferred 
to a higher frequency, which will radiate efficiently through space. 

AM involves simple equipment but is susceptible to noise and interference. 
The SSB-AM system is most economical in its use of frequency space. FM 
requires more complex equipment and employs wide frequency bands but 
noise interference does not limit its usefulness. For equally intelligible signals, 
SSB-AM and FM use much less power than does the AM system. Thus there 
is no one system that is better than the others for all purposes but we can say 
that double-sideband AM is generally less useful than the other systems and 
its application is declining. 

The process of frequency translation to different frequency bands is 
employed in receivers as well as in transmitters and requires the development 
of the product of two frequencies. The output always contains more fre- 
quencies that does the input, sometimes a great many more. Tuned circuit 
filters are ordinarily used to separate the wanted frequencies from the ones 
not wanted. 



REVIEW QUESTIONS 



14.1 What are two fundamental methods of modulating a wave of frequency /? 

14.2 What is the process of amplitude modulation? 

14.3 What is the process of FM ? 

14.4 What is PM ? 

14.5 What are the types of pulse modulation? 

14.6 What is the modulation factor in AM ? 

14.7 Sketch an AM wave as modulated by a square wave. 

14.8 Sketch an FM wave as modulated by a square wave. 

14.9 Why is the bandwidth of an AM signal twice the highest frequency present 
in the modulating signals? 

14.10 What are upper and lower sidebands of an AM wave? How are their fre- 
quencies related to the modulating signal? 



350 Modulation and Detection 

14.11 What is the relation of carrier power to sideband power at 50 per cent modu- 
lation? At 100 per cent modulation? 

14.12 Why do wc filter the output of a diode modulator? 

14.13 At m a = 1.0, what carrier power must be supplied if the modulator output 
is 750 W? What is the power in the upper sideband? 

14.14 A broadcast station is assigned a carrier frequency of 990 kHz and a band- 
width of 10 kHz. What range of frequencies can it transmit in the sidebands? 

14.15 What is per cent modulation in AM? 

14.16 What is meant by SSB? By DSB? 

14.17 For the same power in the information signal, compare the total power in 
AM-DSB and in SSB. 

14.18 How is an SSB signal generated? 

14.19 Compare DSB-AM and SSB on the basis of bandwidth. 

14.20 Explain the operation of a diode detector. 

14.21 What is frequency translation? 

14.22 Explain a product detector. When is it used? 

14.23 What is the most serious problem in SSB reception? 

14.24 What is meant by AVC? Describe AVC circuit performance. 

14.25 What is delayed AVC and why is it used? 

14.26 What is the purpose of a balanced modulator? 

14.27 What is frequency deviation in FM ? 

14.28 Compare AM and FM on bandwidth needs. 

14.29 Why does the power in an FM wave not vary? 

14.30 What is the approximate bandwidth of an FM signal? 

14.31 An FM signal has a deviation of 90 kHz for a • 10-V signal. What is the value 
of k,l 

14.32 An FM station has a channel from 90.8 to 91 MHz. 

(a) What is its center frequency ? 

(b) What is the maximum permissible deviation ratio for a max modulating 
frequency of 10 kHz? 

14.33 What is a limitcr? Why is it needed in FM reception? 

14.34 Explain one form of limitcr circuit. 

14.35 Explain the action of a discriminator. 

14.36 A diode modulator has input frequencies at 2.75, 2.80, and 2.95 MHz. The 
mixing frequency is 2.345 MHz. What output frequencies arc obtained? 

14.37 Explain how to translate a frequency from 30 MHz to 1.65 MHz; give fre- 
quencies involved in the output and explain their separation. 

14.38 What is narrow-band FM? 



Problems 



361 



14.39 What is wide-band FM ? What is its advantage over AM in a noisy channel ? 

14.40 How can per cent AM be found from an oscilloscope pattern? 

14.41 What is AFC? 

PROBLEMS 

14.1 The carrier of an AM wave is at 25 W. What is the average power in each 
sideband at 40 per cent modulation ? 

14.2 A carrier is generated at 150 W and the average modulator power output is 
45 W. What per cent modulation is possible? 

14.3 An AM wave is stated by 

v = 100[l + 0.20 cos 271(1000)/ + 0.05 cos 2n(3000)r] cos 10 6 / 
State all frequencies present in hertz, and give the per cent modulation for 
each. 

14.4 A modulated amplifier operates at 2250 V and a carrier power input of 500 
W. 

(a) What modulator power output will be required for 80 per cent modula- 
tion? 

(b) What is the ratio needed for the Class B output transformer for the modu- 
lator if the Class B stage needs a 7500-Ji output primary load? 

14.5 An AM transmitter has a carrier or 1000 W. What average output must the 
modulator have to reach 80 per cent modulation? What is the total average 
power required by the wave? 

14.6 In Problem 14.5, what would be the per cent saving in power if the same 
information was transmitted by SSB? 

14.7 An AM carrier is at/, - 10 kHz and is modulated 50 per cent by f„ --= 400 
Hz and 20 per cent by A 800 Hz. Plot a frequency spectrum showing proper 
amplitudes if the carrier is shown at 2.5-cm length. 

14.8 If the carrier in Problem 14.7 is at 100 W, what is the total power in the AM 
wave? 

14.9 The current in a diode modulator is given by 

mA, / =- 10 ( 2.5i; + 1 .Or 2 
The voltage applied to the modulator is 

v = [3.0 cos 271(60,000)/ • 0.2 cos 27t(IOOO)/] 
Calculate the amplitude and frequency of all output current components. 

14.10 A carrier of an AM wave has V c - 2.75 V at the detector. What voltage is 
available for AVC action? If m a - 0.5, find the voltage of the audio signal 
at the detector output. 

14.11 A 100-kHz carrier is amplitude modulated by a 5-kHz signal and the upper 



362 Modulation and Detection 

sideband is transmitted. The receiver mixes a signal of 100.3 kHz. What is 
the frequency of the recovered audio signal? Has distortion occurred? 

14.12 An SSB signal f c +f, is applied to a frequency translator along with 
V, sin 2n/,t. Show that the SSB signal may be translated to a higher frequency 
or a lower frequency without distortion of the side frequencies. 

14.13 A 10,000-Hz modulating signal gives m f = 15 in an FM transmitter. What is 
the minimum desirable bandwidth for the transmitter resonant circuits? 

14.14 In an FM transmitter, when the audio frequency is 400 Hz and the audio 
voltage is 2.5 V, the deviation of frequency is 5.7 kHz. What audio voltage 
will cause a deviation of 10 kHz? 

14.15 If the maximum audio frequency applied to an FM transmitter is 5.0 kHz 
and m f = 20, what is the maximum frequency deviation from 10 MHz? 

14.16 A steady carrier at 4.350 MHz is transmitted adjacent to one on 4.354 MHz, 
modulated at 2000 Hz. What output frequencies arc found in a remote 
receiver? Do you see any problems arising? 

14.17 An FM signal has a maximum frequency deviation of 50 kHz and is modu- 
lated by audio signals up to a maximum of 10 kHz. If a receiver has a band- 
width of 50 kHz, is this adequate? 

14.18 An FM station with/ l(m „, -^ 15kHzandm/ = 7operatesat 10-MHzcenter 
frequency. What should be the — 3-dB bandwidth of a tuned circuit to pass 
the major spectrum components of this signal ? What Q would be required 
of the resonant circuit ? 

14.19 An FM system uses 15 MHz as a center frequency. The modulating signal 
when at 10 kHz generates a maximum frequency deviation of 5 kHz. Find 
the bandwidth to pass the necessary components for good fidelity of the 
received signal. 



15 

Radio Systems 



Radio systems are designed to transmit information and the noise appearing 
along the transmission path is the basic low-signal limit for reception. We 
now have sufficient knowledge of circuits and the basic processes of modu- 
lation and detection to understand the overall design of radio receivers and 
transmitters and can discuss some of the system considerations that include 
noise, bandwidth of signal, power, and method of modulation. 

As in most engineering problems, there is rarely a unique answer to be 
found and trade-offs are necessary. 

15.1 Noise 

Random variations of current in electronic circuits are called noise. External 
noise caused by atmospheric discharges or static limits radio reception of 
weak signals below about 20 MHz. At higher frequencies the external noise 
decreases and random currents internal to the circuits produce a noise or hiss 
and this limits weak signal reception. Circuit noise is predominantly due to 
impacts of electrons and atoms in thermal agitation in the materials of the 
circuits. Each impact sends out a short energy pulse with a very broad 
frequency range. The thermal noise produced by a resistance R (an antenna, 
for example) appears as the voltage of an equivalent noise generator 

K, olM = 7.4 X 10-' VTOW) (15.1) 



353 



364 



Radio Systems 



where 

T= temperature, degrees absolute (°C + 273°) 
R = resistance of the circuit 
BW ^ frequency band (3-dB) included in the noise measurement 

For a resistance of 100 £1 at room temperature (300° absolute), the noise 
voltage is 1.28 //V per MHz of bandwidth. 

Since the noise is dependent on temperature, when receiving very weak 
space signals the input stages are sometimes immersed in a case at the 
temperature of liquid helium. This reduces T in Eq. 15.1 and therefore the 
circuit noise. A weaker signal can then be received. 

The amount by which the signal power overrides the noise power in the 
circuit determines the understandability of the signal and we speak of 
signal-to-noise ratio in evaluating circuit performance. The noise contributed 
by an amplifier is measured by the noise figure (N.F.), stated as the ratio of 
the signal power to noise power at the input, SJ N„ to the same ratio at the 
output, SJN„. That is, 

KF.-!fjg (15.2) 

The measurement is usually expressed in decibels. The amplifier noise added 
in the circuit degrades the signal, making SJN less than S,/N,. Some tran- 
sistors arc less noisy than others and less noise is usually produced when the 
transistor is operated at low currents and voltages. Noise figure for input 
stages at high frequencies is usually in the range of 4 to 6 dB, although 2 to 
3 dB can be obtained by careful transistor selection. 

15.2 Information in Signals 

We now need a measurement of the information contained in a signal. 

In a pulsed data transmission system we need to recognize only two levels 
of output per pulse interval: on and off. The quantity of information is 
assumed to depend on the number of such levels or conditions that we must 
recognize in a unit signal; that is, 

/„ = log 2 L bits (15.3) 

where /„ is the number of Ms (basic units of signal), L is the number of recog- 
nizable levels or signal conditions, and the logarithmic base of 2 results from 
the fact that each signal level has an even probability of being recorded 
correctly or incorrectly. Admittedly this is an arbitrary definition but it leads 
to useful and comparative results. 

The single pulse is our unit of information content since with L ^ 2, on 
and off, then 

/„ = log 2 2=1 bit 






Information Capacity ot a Channel 35S 

If we transmit 500,000 pulses per second, each I //s long, the rate of informa- 
tion is high, as 

R - 500,000 bits/s 

With a pulse length of 1 /is, the bandwidth must be I MHz for reasonable 
accuracy of waveform in this data system. 

Speech is our most usual form of information transmittal. With speech we 
can encompass a range of 1000 : 1, or 30 dB, between the weakest intelligible 
sound and the loudest sound in the human voice. It takes an intensity 
change of about 2.5 dB before the ear can notice a change of intensity; with 
an intensity range of 30 dB, we are able to recognize 12 significant levels of 
speech intensity. Thus L = 12 for speech. 

While speech contains frequencies up to about 6000 Hz, good intelligibility 
is possible with the channel narrowed to 3000 Hz, as in the telephone. The 
basic interval in speech is considered to be a half cycle of the highest frequency 
present, or 6000 intervals per second for a 3000-Hz speech transmission. 
With 12 recognizable intensity levels in speech, the maximum information 
rate in speech is 

R = 6000 log 2 12 = 2000 x 3.58 = 20,500 bits/s 

Channel bandwidth is, of course, 3000 Hz. 

A television signal can be similarly analyzed. The difference between white 
and black on a television screen is not very great and the eye can only 
distinguish about 10 levels of gray between the black and white limits. The 
picture is formed in dots on horizontal lines and the eye can resolve about 
500 dots of white, black, or color per line. In the United States television 
system there are 525 lines per picture and 30 pictures or frames per second. 
Putting this data together, we can calculate the information rate in a television 

signal as 

R = 500 X 525 x 30 log 2 10 = 2.62 x 10 7 bits/s 

We see that the speech rate of 20,500 bits/s is low but only requires a 
bandwidth of 3000 Hz. Data pulses transmit information at 500,000 bits/s or 
more and require a few megahertz of bandwidth. Television transmits infor- 
mation at very high rates and utilizes 4.5 MHz of bandwidth. 

Thus we conclude that the bandwidth occupied by a signal is proportional 
to the rate at which information is transmitted. 



15.3 Information Capacity of a Channel 

We now need to measure the information capacity of a transmission channel. 
The more levels we try to separate, the smaller are the differences between 
signal levels and the lower is the accuracy of discrimination in the presence of 



356 



Radio Systems 



noise. A quiet channel will have a greater information capacity than a noisy 
one and to measure the information capacity of a channel we use the signal- 
to-noise ratio. 

Suppose we have a given bandwidth and a received signal power S and 
noise power A'. The power input to the receiver is S + N in the presence of 
N units of noise power. The number of distinguishable levels of signal is 
assumed to increase as the ratio of voltages or 



W s -^=V'+# 



(1S.4) 



since a voltage V is proportional to ,/ Power. 

The information received through the channel per basic signal interval is 



/. = log 2A /l+^ = 4-lo gi (l ! --£) 



(15.S) 



The number of basic intervals per second is the rate of sampling, or twice 
the bandwidth. The information rate is 

1 



/? = 2Bx4-log 2 (l-|) 



The total information that can be transmitted over a noisy channel in total 
time T is 



C=5riog 2 (l I -£-) bits 



(15.6) 



4.U 




























































3.5 
















s 


s 


^^— 










1 f\ 














s 


' 














J.U 






























c *> < 






























1 -■■» 




























































J.O 
































i 




























1.5 


/ 




























1.0 


I 





























12 



14 16 



8 10 

x 
Figure IS. I Logarithms to the base 2. 



An AM Transmitter 



357 



This is known as the Hartley-Shannon law. For its use we provide a curve of 
logarithms to the base 2 in Fig. 15.1. 

The received signal may be partially masked by some noise, and the 
received message will be inaccurate. This means that the S/N ratio is low 
and the channel capacity is reduced. 

The channel capacity for accurate reception of a signal can be increased 
by widening the bandwidth B, as by use of FM or by use of a pulse-code 
system. The time T taken to transmit a message can also be increased, 
possibly by sending a message twice or repeating parts of it as is done in some 
data transmission systems. An increase in channel capacity can be accom- 
plished by raising the transmitter power, giving a larger S/N ratio at the 
receiver. This method is expensive, however, because S/N appears in the 
logarithm term and an increase of power by about 10 is needed to make a 
useful improvement. A cheaper means of improving S/N is often found by use 
of a more effective antenna system, using directional gain for the signal and 
having some directional discrimination against the noise sources. 




Figure 15.2 Block diagram for an AM transmitter. 



75.4 An AM Transmitter 



A block diagram of an AM transmitter is shown in Fig. 15.2. A submultiple 
of the output frequency is usually generated by the oscillator since lower- 
frequency crystals are more stable. A buffer amplifier is used to isolate the 
oscillator circuit from variations in the remainder of the transmitter, 
improving the oscillator frequency stability. The oscillator frequency is then 
multiplied by frequency doublers or triplers. These are overbiased amplifiers 



1 



358 



Radio Systems 



with loads luned to twice or three times the input frequency. In addition to 
frequency multiplication, these amplifiers increase the power level sufficiently 
to drive the modulated amplifier. 

An audio amplifier is driven by the microphone signal at frequencies f t 
and raises the power level sufficiently to drive the Class B modulator to full 
output. The output power for the sidebands is transferred through the 
modulation transformer to the modulated amplifier. This transformer has a 
turns ratio that will produce the desired load for the Class B amplifier and 
transfer the power to the secondary load represented by the resistance of the 
modulated amplifier V cc jl c . Power modulation is obtained and the modulated 
signals arc coupled to the antenna through the it network, which is designed 
to transform the large load value of the amplifier to the 50 or 75 fi that the 
antenna represents. At the same time the n network acts as a filter and 
removes extraneous harmonics from the amplifier output. 

An AM transmitter has simple circuits and is easy to adjust for modula- 
tion without distortion. Amplitude-modulated signals use a carrier that 
transmits no useful information, however, and can be considered to be a 
waste of power; AM signals also require a bandwidth that is twice the modu- 
lating signal band. More importantly, perhaps, AM suffers from heterodyne 
interference, created when another carrier is within a few kilohertz so that 
their difference or beat frequency produces a steady audio whistle in the 
output of the receiver. 

The use of AM is declining because of heterodyne interference, its band- 
width requirements, sensitivity to noise, and its power requirement. 



15.5 The Superheterodyne Receiver 

Reception of AM signals is done with the superheterodyne receiver, shown in 
block form in Fig. 15.3(a). 

The incoming signal has a carrier frequency/., with sidebands at/ +/, 
and/. — /„ where/, is the audio or information signal. This AM signal is 
translated to a new frequency band, with carrier frequency f k and sidebands 
at/* i/„ by mixing with the first oscillator. The oscillator frequency is/ x 
and that frequency is adjusted so that/, = /,+/* for all incoming signals to 
which the receiver is tuned; that is,/, is made to vary in step with /. Ampli- 
fication follows at/*, the intermediate frequency (I.F.), in an amplifier with 
fixed tuning at/*. Since the intermediate frequency is usually lower than the 
frequency of the input signal, the tuned circuits in the l.F. amplifier can have 
more selectivity (in terms of bandwidth in hertz) than can be obtained with 
simple parallel-resonant circuits at the incoming radio frequency. The l.F. 
circuits employ double-tuned transformers that give a flat top and steep 



The Superheterodyne Receiver 



7 

Signal Ciain 

Image Suppn»> 


sion 
i 




' 








RF. 
Amplifier 




Translator 










" 


ft*f, 




u 




Oscillator 



Channel 
Selectivity 


\ 




l.F. 
Amplifier 







Detector 



h f, 



f, 



Audio 
Amplifier 



(a) 



Power 
Amplifier 







Ik'al 
Oscillator 








h = 


h 


l.F. 
Amplifier 


Product 
Detector 




h L h 






« 








Audio 




Power 
Amplifier 




Amplifier 





4 



(b) 



Figure 15.3 Block diagram of a superheterodyne receiver for AM signals; 
(b) demodulation of SSB-AM signals. 

skirts to the overall response curve. This flat response is made just wide 
enough to include all the sideband frequencies present in/* -| /, and/* — /,. 

After l.F. amplification the signal goes to the diode detector and the 
original modulation frequency /, is derived. Audio frequency and power 
amplification follow before the signal reaches the loudspeaker. 

There will always be unintentional positive feedback, introduced by 
common couplings in power supplies, by fields between connecting wires, or 
by stray capacitances. This positive feedback, although small, will limit the 
amount of gain possible at one frequency because when A becomes large, the 



360 



Radio Systems 



product Af) will approach — 1 and instability will occur. By shifting the 
frequency band it is possible to approach the limiting stable gain in each 
frequency band in succession. Thus high gain is an advantage of the super- 
heterodyne receiver. 

Because of the requirement of the superheterodyne system that the 
intermediate frequency f k be a constant for all received frequencies, the 
oscillator frequency /, must continuously differ from the frequency of the 
carrier/, by the amount of the I.F., i.e.,/ t . For any oscillator frequency, two 
signals can give an output to the intermediate frequency amplifier. These 
signals differ by 2f k , one being below/, at/, —f k ^f Cl and one above/, at 
/ w | f k =/ ci . if the desired signal is at/ ci = 545 kHz and the I.F. is/,, = 
455 kHz, then the oscillator must be tuned to/, =/., +f k = 1000 kHz. A 
second signal at f x | f„ = 1000 + 455 = 1455 kHz =/„ can also be 
received simultaneously and would be called the image frequency. 

By using sufficiently selective radio-frequency circuits in the R.F. ampli- 
fier, the strength of the undesired image signals can be lowered and it is 
possible to reduce greatly or to eliminate image responses. At signal fre- 
quencies of 20 MHz or more, however, an I.F. of 455 kHz places the image at 
only 20.910 MHz. The radio-frequency tuned circuits cannot give sufficient 
attenuation for a frequency so close to the desired carrier and images will be 
received with the desired signals. 

To eliminate images in such high-frequency receivers, a higher I.F. is used, 
possibly 5.5 MHz, so that for the 20-MHz signal the image frequency will be 
at 31 MHz and sufficiently separated from the desired signal to be rejected 
by the input tuned circuits. A second translation from 5.5 MHz to 455 kHz 
can be carried out so that the skirt selectivity of the 455-kHz I.F. amplifier is 
also obtained for the 20-MHz received signal. 

Thus the selectivity curve of the receiver is really that of the lower- 
frequency I.F. amplifier. This makes the receiver selectivity curve independent 
of the frequency range for which the receiver is designed. This is a second 
advantage of the superheterodyne form of receiver circuit. 

Single-sideband signals can be received in the same circuit if a second 
oscillator is added to reinsert the carrier frequency as shown in Fig. 15.3(b). 
If the I.F. is 455 kHz, then/, of the second oscillator should also be 455 kHz. 
A product detector is used and the low or difference output frequency 
chosen, giving an output frequency of 455 ±f t — 455 =/.. The plus or 
minus sign is used, depending on whether the upper or lower sideband is 
being received. The output of the detector is /,, the original modulating 
signal. The oscillator frequency f y must be stable within 10 or 20 Hz if 
intelligible speech is to be recovered. 

A receiver for the broadcast band is designed to tune over the frequency 
range from 600 to 1 600 kHz. This is a frequency range of 1 600/600 = 2.67 : 1 . 
The resonant frequency of a tuned circuit varies with \\«f~C for a constant 



The Superheterodyne Receiver 



361 



inductance L. Most variable air capacitors have a maximum capacitance 
about 10 times their minimum capacitance so that «f\Q = 3.16 : 1. This is 
about the usual tuning range for a variable capacitor. The broadcast band, 
requiring a tuning range of 2.67 : 1, can be satisfactorily covered with such a 
capacitor. A small variable capacitance is used in parallel to adjust the 
minimum capacitance and to obtain the exact ratio of maximum to minimum 
capacitance needed for the tuning range. 

For a superheterodyne receiver with the I.F at 455 kHz, the oscillator 
frequency must range between 



Signal Frequency 

600 kHz 
1600 kHz 



Oscillator Frequency 



145 or 1055 kHz 
1145 or 2055 kHz 



The required tuning ratio for the oscillator tuning capacitor would be 
cither 

1145 = 7.9:1 or |°J= 1.94:1 



145 



1055 



The usual air variable capacitor has a range from 35 to 350 pF, or 10 to 1. 
Such a variable capacitor could not tune over the wide-frequency range from 
145 to 1 145 kHz. Therefore the oscillator is designed to tune from 1055 to 
2055 kHz and the oscillator frequency is placed above the signal frequency, 
or/, =/ e +/». Because the oscillator tuning range will be only 1.94: 1 
compared to 2.67 : 1 for the radio-frequency signal circuits, the oscillator 
tuning capacitor usually has a lower maximum capacitance. 

Since the signal frequency is the difference between the oscillator fre- 
quency and the I.F.,/, =/„—/*, the image frequency occurs at a signal 
frequency that is the sum of the oscillator frequency and the I.F., /„ = 
/, -\-f k , and we have 



Signal Frequency 



600 kHz 
1600 kHz 



Oscillator Frequency 



1055 kHz 
2055 kHz 



Image Frequency 



1510 kHz 
2510 kHz 



This places the range of possible image frequencies between 1510 and 2510 
kHz, almost entirely outside the broadcast band and in a region where 
strong signals are not prevalent. Therefore the reception of an image signal is 
not usual, although a strong broadcast station near 1500 kHz may appear 
at tuned frequencies near 600 kHz. 



382 Radio Systems 

15.6 The SSB Transmitter 

A transmitter for SSB signals is more complex than the AM unit and one is 
shown in block form in Fig. I5.4. A radio frequency of perhaps 5 MHz,/,, 
and the audio signal/, are introduced to a balanced modulator and sideband 
pairs obtained without a carrier. A filter selects the upper or lower sideband 
to be transmitted and in a second translator this sideband signal is added to a 
carrier/, such that/, +/,+/»=/, +/,. or/, -/,+/»=/,-/, for the 
lower sideband. This gives a single-sideband signal at the frequency/, at 
which it will be transmitted. The Class B linear amplifier provides a power 
gain for this varying amplitude signal. The output of the amplifier is coupled 
to the antenna through the n network for harmonic filtering of the Class B 
output. 

A frequency of about 5 MHz is chosen for the first oscillator so that the 
sideband filler can adequately separate the upper and lower sidebands, which 
differ by only 300 or 400 Hz. Such filters are usually composed of quartz 
piezoelectric crystals with high-£? values for sharp cutoff. 

/, 



Audio 
Amplifier 



/. f, 



fa+f, 

or 



/c+/x 



Balanced 
Modulator 



Filter 



Mixer 



Class B 
Linear Amplifier 



[/, = 5 MHz 



First 

Oscillator 



',. 




Second 
Oscillator 



Figure 15.4 Block diagram of an SSB transmitter. 

Single-sideband signals are being generally used for point-to-point 
telephone service, especially for frequency multiplex as in the radio relay 
service. Frequency multiplex is illustrated in Fig. 15.5, where eight separate 
speech channels are used to prepare eight lower sideband signals with 
carriers spaced every 5 kHz from 15 to 50 kHz. After being "stacked up'" in 
frequency as shown, the composite signal, with frequencies from 12 to 
49.7 kHz, is used as a modulating signal for a single carrier at 4000 MHz to 
yield a signal capable of transmission by radio relay. 

At the receiver the composite signal is detected back to the original range 
of 12 to 49.7 kHz and the eight signals separated by filters. The proper 



An SSB Transceiver 



363 



300-3000 11/. 
Speech § 1 



Balanced 
Modulator 



I 2.000-14,700 Hz 



15.300-18.000 Hz 



Sideband 
Filler 



12.000-14.700 Hz 



Oscillator 
15.000 Hz 



300-3000 Hz 
Speech # 2 



Balanced 
Modulator 


17.000-19.700 Hz 


Sidebanc 




20,300-23,000 Hz 


Filter 


! 






Oscillator 
20.000 Hz 






r- 



I 7,000-19.700 Hz 



Other Similar Signals -J 



22.000-24.700 Hz 
27.000-29.700 Hz 
32,000-34,700 Hz 
37.000-39.700 Hz 
42.000-t4.700 Hz. 
47.000-49.700 Hz 



-One Channel 



Figure 15.5 Frequency-division multiplexing of eight voice channels. 

carriers for each channel are added in separate product detectors and the 
original speech signals recovered. 



15.7 An SSB Transceiver 

The block diagram of Fig. 1 5.6 illustrates a combined transmitter-receiver or 
transceiver, a type now commonly employed for SSB at high frequencies. The 
same frequency is used for both transmitting and receiving and the transmit- 
ting path of signals is indicated by dashed lines in the diagram. 

Let us assume that we tune to an SSB signal of upper sideband, derived 
from a carrier at 7.2 MHz. The band of frequencies near 7.2 MHz is amplified 
and mixed with a fixed oscillator operating at ll MHz. The difference 
frequencies are in a band near 4 MHz, including the desired signal, now at 






► 









I! 



2: 



T 

i 



c 
§ 

03 

(/i 

V) 

c 

< 



•?. 



L 



■n 



S° 



c/5 



p: 





I 




N 


■a 2 




O 


£1 




™ 




V* 


o 




o- 











u 


(J 


c 


</: 


« 


c 


w 





■a 

a u 
13 -a 



Lu n 



5 o 






o 

2 w» 



.S«5 

If 

< E 
< 







n. 
E 



391 



,4n SSB Transmitter 



365 



II — 7.2 = 3.8 MHz. This band is mixed with the output of a variable- 
frequency oscillator, tunable over a 500-kHz range from 9.5 to 10 MHz. 
When the oscillator is set at 9.8 MHz, our desired signal will produce dif- 
ference frequencies close to 9.8 — 3.8 = 6.0 MHz. Only our sideband signal 
can pass through the filter with a narrow passband of about 2.5 kHz at 
6.0 MHz, as shown in Fig. 15.7. All other signals in the original narrow band 
selected by the radio-frequency tuner arc removed here by the filter. 

An I.F. amplifier provides gain for the 6.0-MHz signal, which is then 
applied to the product detector. Here a 6.0-MHz carrier is reinserted and the 
speech frequencies/, are obtained as output. 

In the transmit mode the speech signal /, passes through the speech 
amplifier to a balanced modulator, where a 6-MHz signal is supplied from 
either of two carrier oscillators. These differ slightly in frequency so that the 
selected sideband will be centered in the passband of the same filter used for 
receiving at 6.0 MHz + 2.5 kHz. The signal at 6 MHz is mixed with the 
variable-frequency oscillator output at 9.8 MHz, giving a 3.8-MHz output. 










r^\ 




-6 

i it 




- ■/*- 


-2.6 kHz- A 




lu 












in 
























e 

I -30 

E 

g 










- 




















1 


tu 










1 


SCi 












3U 












60 




4 


t.9klly 


1 



(..000 



6.001 6.002 

MHz 



6.003 



Figure 15.7 Selectivity characteristics of 6-MHz sideband filter; 60dB,'6dB 
shape factor = 4.9/2.6 = 1.9. 



366 



Radio Systems 



A second mixing with the 1 1 -MHz oscillator frequency gives a difference 
frequency at 7.2 MHz, actually the same frequency as originally received. 

Power amplification with a driver and power amplifier operating in Class 
B follows. 

Advantages of SSB transceiver operation include 

1. Only a single channel is used for transmission and reception. 

2. SSB gives a narrow bandwidth equal to that of the modulating signal. 

3. The transceive mode employs less expensive equipment, especially the 
filter. 

4. The power required is only that of one sideband of an AM signal. 

The complexity of the equipment is the offset to these advantages. 



15.8 AM versus FM 

For an AM signal the bandwidth is fixed. The rate of transmittal of informa- 
tion can be improved by increasing the transmitter power to raise the SjN 
ratio but this is expensive. With an FM signal, the bandwidth can be increased 
arbitrarily by increasing m f , assuming that frequencies are available. We can 
then receive a useful signal with a reduced S/N ratio. 

For instance, the total signal power in a 1 00 per cent modulated AM wave 
might be 150 W, while the power in an FM wave carrying the same informa- 
tion need be only that of one AM sideband or 25 W. The bandwidths might 
be 10 and lOO kHz, respectively. It is found that the information capacity or 
the usability of the FM channel is nine times greater than that of the AM 
channel; in fact, the FM system is superior as long as the 2/, value is greater 
than 10 kHz, the bandwidth of the AM signal. 

Since the power in an FM wave is constant, there need be no provision 
for the large instantaneous power and voltage peaks of an AM transmitter. 
The FM equipment can be smaller and cheaper. 

Noise signals caused by atmospheric static or man-made electrical 
discharges create reception problems for radio signals. In AM these noise 
signals may override the wanted signal and they cannot be separated. The 
noise can be reduced by narrowing the reception bandwidth but the minimum 
bandwidth of AM signals is determined by the modulating frequencies of the 
signal. 

If a receiver is designed for reception of wide-band FM signals and made 
insensitive to amplitude-varying noise by use of a limitcr, the noise can be 
largely eliminated. 

Interference between two signals on the same frequency is a major problem 
in radio communication. In AM, if the two signals have carriers differing 



FM Systems 



367 






only a few kilohcrtz, the difference appears as heterodyne interference, or a 
steady whistle in the receiver output. This is a frequent occurrence in the 
crowded AM bands. 

When this type of interference occurs between the higher-order sidebands 
of two AM broadcast stations with carrier frequencies spaced 10 kHz apart, 
the resultant noise is called "monkey chatter." Filters are often added to AM 
receivers to eliminate this form of interference. 

When two FM signals arc on the same center frequency, we encounter 
the capture effect, by which the stronger signal captures the receiver and 
blocks reception of any signal significantly weaker. Wider frequency devia- 
tion makes this capture effect greater. E.H. Armstrong was the first to 
demonstrate the value of FM in combating interference. 

Thus FM has a number of advantages over AM, particularly where noisy 
channels must be used with weak signals as in the mobile service and where 
power input must be limited. 



15.9 FM Systems 

The FM transmitter diagrammed in Fig. 15.8 employs a variable-capacitance 
diode to shift the frequency of an oscillator in accordance with the modu- 
lation signal. The varying oscillator frequency may be generated with a 
center frequency of/, at which a suitable frequency deviation can be obtained. 
That varying frequency is translated by the output of a second oscillator at 
/», such that/, +f k =/,, where/, is the assigned output frequency of the 
transmitter. The side frequencies are then clustered around /, in the same 
manner that they were originally generated around/,. The power amplifier 
provides the antenna power at/, and side frequencies. 



f, 



I 



t.*h m f. 









Oscillator 


U 


Frequency 

Translator 




Power 


/. 


Uioui" 








Amplifier 














V 














Second 
Oscillator 




















Varying DC 


Frequency 


















Discriminator 





Figure 15.8 A reactance-modulated FM transmitter. 



368 



Radio Systems 



The output is also applied to the tuned circuit of the frequency discrimi- 
nator of Sec. 14.13, which yields a voltage as a measure of the average drift of 
frequency from the assigned channel at/,. If the frequency is not at normal, 
a correction voltage from the discriminator is applied to the frequency- 
controlling diode, to change its capacitance and to restore the oscillator 
frequency to its assigned channel. 

The transmitter is relatively simple because of the constant power require- 
ments of an FM signal. Circuit bandwidths can be adjusted by variation of 
circuit Q to obtain sufficient bandwidth for transmission of the important 
side frequencies. 

The FM receiver of Fig. 15.9 follows the general design of the super- 
heterodyne up to the detector. At that point a limiter-discriminator circuit is 
inserted to change the frequency variations to amplitude variations. The 
audio amplifier following the discriminator is standard. 

To provide for adequately large frequency deviations to reduce inter- 
ference, FM signals are used in the frequency bands above about 40 MHz. 
An I.F. of 10.7 MHz has become standard so that image interference will not 
be serious. The l.F. of 10.7 MHz also permits the design of double-tuned 
transformers with adequate bandwidth to handle the FM deviations. With 
200-kHz radio-frequency bandwidths, some loading of the tuned circuits 
with parallel resistances may be necessary to obtain the needed bandwidths. 



I 



/,. + <V„ 



A + a/„ 



R.F. 
Amplifier 




Mixer 




Amplifier 




Limiler 




Discriminator 








( 


f. 














h 






Oscillator 




Audio 
Amplifier 


-4 



Figure 15.9 An FM receiver. 



15. 10 A Radar System 

The radar system of Fig. 15.10 illustrates the manner in which electronic 
systems are designed by the assembly of relatively simple component circuits 
and functions into an ultimate complex objective. , 

A typical radar system employs a pulse sequence as in Fig. 15.11(a), 
transmitting a short very powerful pulse at a rate of perhaps 400 per second. 
This radio signal travels to a distant target and a very small portion of the 
energy is reflected back to the receiver. The time taken for the round trip is 







369 






370 



Radio Systems 



Antenna 
Rotation 



2.5 ms- 




CRO Screen 

<"> (b) 

Figure 15.11 (a) Transmitted radar pulses ; (b) received radar signal as presented 
on the CRO plot. 

a measure of distance to the target and is plotted as radial distance on a 
cathode-ray oscilloscope screen with a linear radial sweep, rotating in step 
with the rotating antenna. The returned pulse is used to brighten the scope 
display at its instant of arrival, presenting a visual dot at the target position, 
as in Fig. 15.11(b). 

The transmitter starts with a pulse generator, at 400 Hz, followed by a 
limiter producing square waves. These are used to produce short pulses at 
400 times per second. An oscillator at 30 MHz drives modulator I, which is 
turned on by the pulses as in Fig. 15.1 1(a), and results in pulses of 30 Mhz 
signal with pulse lengths of a fraction of a microsecond. By mixing with 
another oscillator at 270 MHz, the final pulses are obtained at 300 M Hz. These 
are amplified to hundreds of kilowatts and radiated in a very narrow beam 
by a sharply directive antenna. 

While the pulse is transmitted, the receiver input is blocked to avoid 
overload. The receiver input is opened after the pulse is sent to await the 
returned signal or echo. The echo pulse will be at 300 MHz ±A, where A 
represents a shift in frequency if the target is in motion. This is the Doppler 
effect and A = Ivfjc, where v is the target velocity and c is 300 X 10«, the 
velocity of radio waves in meters per second. 

The received echo is amplified at 300 MHz and then mixed with the 
270-MHz oscillator frequency in the first translator. The signal is there 
converted to 30 MHz ± A; here wide-band amplification is used to preserve 
the pulse waveform. The signal is then mixed with 30 MHz from oscillator I. 
Retained in the output is the pulse envelope at 400 repetitions per secpnd and 
this is passed to the cathode-ray oscilloscope to control the brightness of the 
spot on the screen. 

The delay gate can be used to open the detector circuit at a given time after 
the pulse is transmitted so as to receive only a single returned pulse. In 



Review Questions 



371 



automobile speed measurement on the highway, this gating allows the 
observer to concentrate on the speed of only one target. 

The basic circuit form of the receiver is that of the superheterodyne. To 
the equipment above must be added numerous power supplies and control 
equipment for feeding the antenna position into the oscilloscope. While 
rather easy to describe, a radar is, in fact, a complex system. 



15. 1 1 Frequency Classification for 

Radio Signals 

Portions of the broad radio spectrum have varying characteristics and in 
describing radio waves and the equipment suited to them, we have the broad 
classification of Table 1 5. 1. 



TABLE ISA Radio-Frequency Classification 



Hand Classification 



Frequency Range 



Use 



Low frequency (LF) 

Medium frequency (MF) 
High frequency (HF) 

Very-high frequency (VHF) 

Ultra-high frequency (UHF) 
Exlremc-high frequency (KHF) 



30-300 kHz 

300kHz-3MHz 
3-30 MHz 

30-300 MHz 

300 MHz-3 GHz 
3-300 GHz 



Marine point-to-point; 
navigation systems 

Commercial broadcast 

Moderate and long-distance 
communications 

Television, FM, aircraft 
navigation 

Television, radar 

Radar, space communica- 
tions, radio relay 



REVIEW QUESTIONS 

15.1 What is meant by the signal-to-noisc ratio at the receiver input? 

15.2 Why do receivers for signals from space use input circuits cooled to liquid- 
air temperatures? 

15.3 How does noise vary with bandwidth received? 

15.4 What is meant by the noise figure of a receiver? 

15.5 What is thermal noise? 

15.6 What is a bit of information? 

15.7 In computing information content of a signal, why is the logarithm taken to 
the base 2? 



372 



Radio Systems 



15.8 You are receiving an important telephone conversation and ask the caller to 
repeat a word. What have you done to the channel capacity? 

15.9 What are the three factors available for trade-off in increasing the bit capacity 
of a communications channel ? 

15.10 Why is it uneconomic to increase power to improve the channel capacity? 

15.1 1 What is the cause of heterodyne interference in AM reception ? 

15.12 What determines the bandwidth of an AM signal? 

15.13 What situation leads to a steady whistle in the output of an AM receiver? 

15.14 List the major circuit functions employed in a superheterodyne receiver. 

15.15 Name two major advantages of the superheterodyne receiver. 

15.16 What is meant by I.F.? 

15.17 What is the reason that we can obtain very high signal gain in a superhetero- 
dyne receiver? 

15.18 What is an image frequency? 

15.19 A signal at 970 kHz is being received with an oscillator at 1430 kHz in a 
superheterodyne receiver. What is the I.F.? What will be the frequency that 
might be received as an image signal? 

15.20 Name two advantages in the use of SSB over double-sideband AM. 

15.21 Why must the carrier reinsertion oscillator of an SSB superheterodyne receiver 
be very stable? What are its stability limits in frequency? 

15.22 What is a transceiver? 

15.23 Can you see an advantage for the transceiver form of SSB transmitter and 
receiver? 

15.24 Name several advantages of FM over AM. 

15.25 What is the capture effect in FM reception? 

15.26 From the Hartley-Shannon law, explain why increasing the bandwidth 
improves FM reception. 

15.27 What causes the Doppler effect in a received radar signal? 

15.28 The time from transmitted pulse to echo reception in a radar set is 30.5 /is. 
How far away is the target ? 

15.29 What frequency range is covered by VHF signals? 

15.30 What is a major service employing VHF frequencies? 

15.31 What frequency range is covered by UHF frequencies? 



PROBLEMS 

15.1 A teletypewriter uses a code with five possible positions for holes in a paper 
tape per letter symbol. Letters and spaces are sent at the rate of 360 per 
minute. What is the information rate in bits per second? 



Problems 



373 



15.2 A radio receiver has an input resistance of 70 £2. A signal of 4.7 //V is applied 
along with a steady noise voltage of 0.97 /zV. What is the signal-to-noise 
ratio at the input? 

15.3 The receiver of Problem 15.2 has a noise figure of 3.7 dB. What is the signal- 
to-noise ratio in dB at the receiver output? 

15.4 How many bits can be transmitted per second through the channel repre- 
sented in Problem 15.2 with a 3-dB bandwidth of I0 5 Hz? 

15.5 The signal received in Problem 15.2 is reduced to 1.7 fiV across the 70-S2 
input circuit. With a 3-dB bandwidth of 10 s Hz and circuit temperature of 
300°C absolute, what is the signal-lo-noise ratio at the input? 

15.6 A wirephoto picture size is 12.5 cm x 18 cm and it is scanned at a rate of 40 
lines per centimeter, with an equal resolution or number of dots along the 
lines. We assume that the eye can see 10 different levels or density gradations 
in the picture. Determine the SIN ratio in decibels required for the channel, 
of bandwidth 1200 Hz, if the picture is scanned in 3 minutes. 

15.7 A superheterodyne receiver is tuned to a signal at 1.85 MHz. A signal at 
2.76 MHz is found to interfere. What is the reason? What is the I.F. of the 
receiver? What is the receiver oscillator frequency? 

15.8 A superheterodyne receiver tuned to a signal at 21.7 MHz has its oscillator 
at 25.2 MHz. What is the l.F. of the receiver? What is the possible image 
frequency? 






16 

Digital Circuits 



Digital signals, using pulses, have become an increasingly important part of 
electronics in recent years, with the widening application of data processing 
and the digital computer. Beneath the complexity of these devices we find 
relatively simple and inexpensive circuits. This simplicity results from the 
fact that the circuits employ only two current levels or slates, ON and OFF, 
in a binary or two-level code. 

For a transistor to discriminate among 10 levels of current for decimal 
numbers would place an almost impossible requirement on transistor ac- 
curacy. But electronic devices are well suited to a binary code of on and off 
pulse signals, which require only that the transistors operate at saturation 
(ON) or cutoff (OFF). The signal-present slate is usually designated 1 and 
the off state as 0; thus binary or two-level numbers are represented by chains 
of on-ofl" pulses or l's and 0's. 

There are two basic types of circuits used in handling binary signals. They 
are logic on-off gates and multivibrator switches. We shall show how these are 
combined to process digital signals. 

16.1 Binary Numbers 

The decimal number system, with a base or radix of 10, has been a part of our 
lives since early childhood. With the decimal system we emnloy 10 marks or 
symbols, which wc designate as 0, 1, 2, 3, .... 9. 



374 



Binary Numbers 



376 



If we explore the meaning of position in decimal numbers, wc shall sec 
that a numbering system with a base other than 10 is quite possible; wc might 
use 2, 3, 8, or 9 as examples. Each position or place in a decimal number car- 
ries with it a weighting factor expressed in powers of 10: 



Power of 10: 10« 



105 



10* 



10 J 10* 10' 10° Decimal 
point 
Weight of 1.000,000 100,000 10,000 1000 100 10 1 
position: 

We use a zero to indicate that a particular weighted value is absent; we 
use our numbers, 1, 2, 3, . . . , 9 to tell us how many of a particular power of 
10 are present at a position. 

For instance, 521 in the decimal system employs the first three positions 
to the left of the decimal point and may be expressed 

521 = (5 X 10 2 ) + (2X 10') + (1 X 10°) 
= 500 + 20 + 1 
The position of a digit is weighted by 10 2 = 100, 10' - 10, 10° = 1, respec- 
tively. 

In the binary system of numbers, the idea of position is the same as in the 
decimal system but powers of 2 are used for the weights with each position. 
Only two symbols are used, 1 and 0. The weighting factors are 



Power of 2 2» V 2« 2' 2* 2> 2"- 2' 2° Binary 

point 
Weight of 256 128 64 32 16 8 4 2 1 
position 
(decimal 
value) 

The decimal number 325 is expressed as 101000101 in the binary system. 
This means 

(1 x 2») -(Ox V) + (1 x 2») + (0 x 2 s ) + (0 X 2*) 

+ (0 x 2 J ) + (1 x 2 2 ) + (0 x 2') I- (1 x 2°) 

In decimal values we have 256 + 64 + 4 + I, which totals 325. The pre- 
viously used decimal 521 would be written 1000001001. 

In the decimal system, a decimal fraction such as 0.812 is expressed as 

0.812 = (8 x 10"') -f (I x 10" 2 ) + (2 x 10" 3 ) 

Similarly the binary fraction 0.1 101 means 

0.1 101 = (I x 2-") + (1 X 2" 2 ) |- (0 X 2" 3 ) + (1 X 2-) 



376 



Digital Circuits 



Giving this the weight in decimal numbers 

= 0.500 + 0.250 + 0.062 = 0.812 (decimal) 

A simple method for conversion of a decimal number to binary represen- 
tation employs repeated division of the number by 2. The remainder of I or 
after each division becomes a digit of the binary number. For the decimal 
number 232 we have 



Least significant digit 





Remainder 


232 H- 2 = 116 





116 -r 2= 58 





58 ■¥ 2 = 29 





29 + 2 = 14 


1 


14 -f 2 = 7 





7 -:- 2 = 3 


1 


3 -r 2= 1 


1 


1-4-2= 


1 



Most significant digit 



The last 1 obtained is the largest digit in the binary number. For decimal 232 
we have 1 1 101000 as the binary expression. 

Fractional decimal numbers may be converted to binary form by succes- 
sive multiplications by 2. For each step that results in a 1 to the left of the 
decimal point, record a binary 1 and carry on with the fractional portion of 
the decimal number. With a to the left of the decimal point, record a binary 
and carry on. For instance, to convert decimal 0.9375 to binary form, we 
operate as follows: 





Binary 


0.9375 x 2 = 1.8750 


1 


0.8750 x 2 = 1.7500 


1 


0.7500 x 2 = 1.5000 


1 


0.5000 X 2 = 1.0000 


1 


0.0000 x 2 = 0.0000 






Most significant digit 



Least significant digit 






The binary equivalent of decimal 0.9375 is expressed as 0.1 1 1 10. The largest 
digit is the first binary number obtained and it is placed to the right of the 
binary point. 

The requirement for only two symbols in binary simplifies our electronic 
circuits but we offset this simplicity with the necessity for handling many 
more digits in the binary representation. For instance, decimal 10 is 1010 in 
binary. 

A digit in binary is referred to as a bit, from the initial and final letters of 
binary digit. 



Binary Arithmetic 



377 



16.2 Binary Arithmetic 

Addition and subtraction in binary numbers is easier than the procedures 
used in the decimal system. Four rules apply for addition and these can be 
stated: 

+ 0-0 

0+1-1 

1+0=1 

1 + 1=0 with a forward carry of I 

A carry is handled similarly to decimal system procedure. Binary addition 
can be demonstrated in the following: 



Binary 




Decimal 


1 101 1 




27 


1011 




11 


10000 


Sum 




1 11 


Carry 




00110 


Sum 




I 


Carry 




100110 




38 



Subtraction in digital computers is almost universally handled by chang- 
ing the negative number to its complement and adding. Negative numbers are 
usually manipulated in complement form. 

With decimal numbers we form the 9's complement by subtracting each 
digit of the negative number from 9. We then add 1 to form the 10's com- 
plement. Suppose we wish to subtract 548 from 2012. The 9's complement of 
548 is 451 and 451 + 1 = 452, which is the 10's complement of 548. To 
subtract, we add the 10's complement and drop 1 from the leftmost digit: 



Negative Numbers 


Complementary 
Subtraction 


2012 

- 548 

1464 


2012 
+ 452 

2464 
-I 




1464 



In binary numbers we employ a similar process. To form a l's comple- 
ment, we subtract each digit of the binary number from I. For instance, the 



378 



Digital Circuits 



l's complement of 1001 1 is 01 100; to change this to the 2's complement, we 
add I so that 01 101 is the 2's complement. To subtract 1001 1 from 1 1001, 
as an example, we add the 2's complement: 



2's complement 



11001 

01101 

100110 

1 



00110 

after subtracting 1 from the leftmost digit. 

Since 1 1001 — decimal 25 and 1001 1 - decimal 19, the result should be 
25 - 19 = 6; in binary, 001 10 = decimal 6. 

In the decimal multiplication process, we add the multiplicand to itself 
the number of times specified by the least significant term in the multiplier, 
shifting one place to the left and repeating for each term of the multiplier. 
Binary multiplication follows the same procedure but is simplified because 
each term needs to be set down only once to represent multiplication by 
binary I. 

Thus 10101 x 1011 (decimal 21 x II) leads to 

10101 
101 1 



10101 
10101 
00000 
10101 
11 1001 1 1 



Multiply by I 
Shift, multiply by I 
Shift, multiply by 
Shift, multiply by 1 

Sum — decimal 231 



Binary division is carried out by successive subtractions or additions us- 
ing the 2's complement in each case plus shifts to the right. 

Thus, with the use of complements, all four of our arithmetic operations 
are reduced to the process of addition, plus right or left shifts. This greatly 
simplifies the arithmetic unit of a computer. 



16.3 Binary Codes 

To convert from the decimal system of the business community to the binary 
system in the digital computer has called for the development of computer 
codes. Foremost among these is the binary-coded-decimal (BCD) representa- 
tion. The BCD code uses four binary bits to represent each of the 10 decimal 
digits. This form is also called the 842 1 code, a name derived from the weight 
of each the four binary digits. The code groups are simply derived from binary 
representations as shown in Table 16.1. 



Other Number Systems 



379 



TABLE 16.1 The BCD Code 



Decimal 


Binary 


Decimal 


Binary 





0000 


5 


0101 


1 


0001 


6 


0110 


2 


0010 


7 


01 1 1 


3 


0011 


8 


1000 


4 


0100 


9 


1001 



The decimal number 3582 would be represented in four code groups as 
0011 0101 1000 0010 

Binary-coded-decimal form is frequently used at computer input and 
output. Switching circuits are available to convert such groups of bits to 
full binary number representation or to decode BCD groups into decimal 
numbers. Such circuits are usual at the output of most digital-reading 
instruments, such as digital voltmeters. 

The number of bits employed in a code increases as does the number of 
symbols represented; that is, three bits are sufficient to transmit the eight 
decimal numbers 0, . . . , 7, while four bits are needed to transmit decimal 
numbers 0, . . . , 15. To transmit the English alphabet of 26 letters requires 
that 2" > 26 bits be used; with V = 32, a five-bit code allows 32 difierent 
characters to be represented and the letters, punctuation marks, and start- 
stop signals can be transmitted. This is the alphanumeric code used for 

teletypewriters. AC /-mi 

There is also in common use a code with seven bits, known as ASCII 
(American Standard Code for Information Interchange). With V =-■ 128 com- 
binations of bits, the numerals, capital letters, and lowercase letters are 
available. Actually there are 64 characters and 64 operational controls. An 
example of the latter is the group 0001 010, which calls for a line feed or the 
movement of the printing position to the next line. The computer can readily 
differentiate between a control function with the first two bits as zeros and 
an alphanumeric character when the first two bits are not both zeros. 

Binary digits are sometimes handled in groups called bytes. The number of 
bits in a byte is not standardized but in one system a byte consists of eight 
binary bits, which might include two decimal digits in BCD code or one 
alphanumeric character in an eight-bit code. 

16.4 Other Number Systems 

Another number system employed in computers uses the base 8 and is called 
the octal system . By processing base-8 numbers in the binary-coded form, we 
need employ only the 1,0 representation or the on-ofl" switching of binary 



380 



Digital Circuits 



numbers. Therefore we can use ihe base of 8 without requiring our electronic 
devices to recognize more than the usual on-off states. 

The number positions are given weights of powers of 8 and the symbols 
employed are 0, 1, 2, . . . , 7. The number 352 in octal means 

octal 352 = (3 x 8 2 ) |- (5 x 8') + (2 x 8°) 
= (3 x 64) 4- (5 X 8) +(2 X I) 
= 192 + 40 + 2 = 234 decimal 

To transmit octal 352 in binary-code groups we use 

0011 0101 0010 

Since 7 is the largest number to be represented in octal, we could reduce this 
to 

011 101 010 

Note that the binary system representation for decimal 234 is 

OTTTOTOTO 

so that if a binary number is segmented into groups of three bits, we have the 
binary code for octal numbers. This easy conversion from binary numbers to 










TABLE 16.2 


Number Systems 


Decimal 


Binary 


Octal 


Hexadecimal 














1 


1 


1 


1 


2 


10 


2 


2 


3 


11 


3 


3 


4 


100 


4 


4 


5 


101 


5 


5 


6 


no 


6 


6 


7 


Ml 


7 


7 


8 


1000 


10 


8 


9 


1001 


II 


9 


10 


1010 


12 


A 


11 


1011 


13 


B 


12 


1100 


14 


C 


13 


1101 


15 


D 


14 


1110 


16 


E 


15 


mi 


17 


F 


16 


10000 


20 


. ,0 


17 


10001 


21 


1 11 


18 


10010 


22 


12 


19 


1001 1 


23 


13 


20 


10100 


24 


14 



Series and Parallel Processing of Bits 



381 



octal binary code is one reason for the employment of octal numbers within a 
given computer. 

The representation of decimal 8 and 9 requires four bits in BCD code. 
Without adding any more equipment, we can expand the BCD representation 
to a base- 16 system, or a hexadecimal system, with 16 symbols employed as 
0, 1,2 9, A, B, C, D, E, and F. Decimal 10 is represented as A. 

Table 16.2 shows the relationship of the several number systems dis- 
cussed. 

Subscripts are used to identify the number system employed. For example, 
352 as an octal number would be written as 352„ ; 01 1 in binary representation 
would be 01 1 2 ; and decimal 654 would appear as 654 l0 . 



16.5 Series and Parallel Processing 

of Bits 

The internal language of the digital computer is in the form of binary num- 
bers and codes. The binary signals arc handled as timed chains of pulses. A 
clock or timer generates the basic frequency, as shown in Fig. 16.1(a), and 
binary switches operating in on and off states code this waveform to represent 
binary data in on (1) and off (0) pulses, as shown in Fig. 16.1(b). The pulse 
chains may actually be positive and zero, zero and negative, or positive and 
negative in amplitude and there are many codes in which the data may be 
transmitted such as BCD or the seven-bit ASCII code. 

There are two basic methods of handling bits in data processing. When 
transmitted over a wire line or a radio channel, the bits arc sent in series and 
operated upon in sequence; the pulses in Fig. 16.1(b) may be visualized as 
sliding off the page to the right, with the first pulse being the most significant 
bit of the number. In parallel operation, the bits in the figure may be thought 
of as sliding down the page simultaneously in four separate channels; the 
weight ascrj^l to each bit is determined by the channel in which the pulse 
appears. 



'Mim? j 













Most 
■*"" Significant 
Bit 

Series 


1 





1 1 


** Output 


Binary 1 101 




111! 






Parallel Output 







(a) (b) 

Figure 16.1 (a) Clock or timing pulses; (b) binary code 1101 = decimal 13. 



T 



382 



Digital Circuits 



For a signal of 40 bits, the time for serial processing is 40 times that for 
the processing of 1 bit. With parallel operation the 40 bits are simultaneously 
processed and the time is approximately that for operating on 1 bit. Thus 
parallel operation would be 40 times as fast but requires 40 times as much 
equipment. 

For reasons of speed, most digital computation is done with parallel 
processing. Registers or storage elements are used to accumulate n bits as 
they are received in series and the storage is periodically emptied or (lumped 
into ii parallel channels. 



16.6 Logic Operations in Addition 

It has been demonstrated that the basic arithmetical operation in digital 
computation is that of addition. To further analyze the addition process, we 
again carry out an addition with binary numbers A and B: 



A 


ion 




B 


Oil I 






[100 

11 


Sum 
Carry, C, 




1010 


Sum 




1 


Carry, C 2 




10010 


Sum 



The requirements for the addition operation are summarized in the table 
below, in which columns A, B, and C, provide for all possible combinations 
of or I input signals for these variables. The columns S and C z arc the 
required results, reasoned from the addition example above and confirmed 
by the addition rules of Sec. 16.2. 



A 


B 


c, 


5 


Ci 























1 


1 








1 





1 








1 


1 





I 


1 








] 





1 





1 





I 


1 


I 








1 


1 


1 


1 


• 1 


1 



The addition will be carried out in parallel channels so that one adding 
circuit will handle one bit from A and one bit from B. The process is not 



Logic Switches 



383 



numerical in nature but requires circuits that provide I- or 0-Icvel output 
signals in response to several I- or 0-levcl command signals at the inputs. 
Specifically, digital operations utilize circuits that recognize the presence of 
a pulse from one circuit and the presence of a pulse from a second circuit, 
as well as when a pulse from one circuit or a pulse from a second circuit is 
present. Also there must be recognition of a carry pulse C,, coming from a 
preceding bit circuit, and provision must be made to transmit a forward 
carry C 2 to the next most significant bit channel. 
The table shows that the sum S equals 1 when 

A = I OR B = I OR C, = I. other variables = 

or (16.1) 

a and b and c, = 1 

These statements are reasoned from lines 2, 3, 5, and 8 of the table. 

We also see that there must be a forward carry output C 2 , or C 2 = 1, 
when 

A and fl= 1. C, =0 

OR 
/JandC, = 1, B = Q 

OR (16.2) 

B AND C, = 1, A -- 

OR 
A AND B AND C, = l 

These statements result from lines 7, 6, 4, and 8 of the table. 

To carry out the addition operation we need circuits that yield proper 
I- or 0-level results from or and and combinations of two or three pulse 
inputs. We shall later find that a not or inverter operation is also needed. An 
inverter gives a 0-level output for a 1 -level input, or vice versa. 

Such circuits that recognize the presence of a group of 1- or 0-level signals 
at the input and give I- or 0-level output dependent on or, and, not com- 
binations of these input orders are called logic circuits. They act as off-on 
switches. 

16.7 Logic Switches 

Logic circuits are often called gales because they are open (OFF) or closed 
(ON) as called for by the specified combinations of pulse inputs. 

A logic and gate must provide a "logic-!" output only if all inputs are at 
the I level. The and operation is illustrated in Fig. 16.2(a) as a series connec- 
tion of switches. Only if all the switches arc closed or at the 1 condition will 
an output voltage appear at F; that is, we write in logic algebra 

ABC = F (16.3) 



384 



Digital Circuits 




I 



Ao- 
Uo- 



(o- 



<a) 



(1)1 



Figure 16.2 (a) Switches in an and circuit; (b) and logic-circuit symbol. 

and read this as A and B and C equals F. The variable F is equal to 1 if the 
switches arc all closed ; F — if any switch is open. 

A table of switch outputs for an and circuit with A and B inputs is 



A 


B 


F 














1 





1 








1 


1 


1 



AND 



A circuit symbol for an and circuit is drawn in Fig. 16.2(b). 

A logic or gate must provide an output at "logic- 1" level if any one or 
more of its inputs is at the 1 condition. The or operation is illustrated in Fig. 
16.3(a) as a parallel connection of switches. If switches A or B or C, or any 
combination, is closed, the circuit is complete and F= I; that is, in logic 
algebra 

A + B+C=F (16.4) 

which is read as "A or B or C equals F." 

A table of switching outputs for an OR circuit is 



A 


B 


F 














1 


1 


1 
1 



1 


1 

1 



A circuit symbol for an or circuit is shown in Fig. 16.3(b). 

The concepts of multiplication or addition should not be associated with 
the indicated symbols in Eq. 16.3 and 16.4; the equations are intended to 
convey ideas of switch connection. 

A logic not circuit supplies the inverse of any operation and the inverse 
of switch A is written A {A not). If A — 1 , then A = 0, and vice versa. A 
common-emitter connected transistor often supplies the not operation with 
its inversion of signal. 






Logic Switches 



385 




► /•' 



Ao- 
«o- 



Cc- 




(a) (b) 

Figure 16.3 (a) Switches in an or circuit; (b) or logic-circuit symbol. 

By combining the and and not operations, we can form a nand (negative 
and) gate. A table of switching operations is 



A 


B 


F 








1 





1 


1 


1 





1 


1 


1 






NAND 



A circuit symbol for a nand operation is shown in Fig. 16.4. 

Similarly, if we combine or and not gates, the result is a nor (negative 
or) operation. An operations table for the nor gate is 



A 


B 


F 








1 





1 





1 








1 


1 






A circuit symbol for a nor gate is drawn in Fig. 16.4. 



/lO- 

fio- 




Ao- 



F = AB Bo- 




F = A+li 



(a) 



(b) 



.-to- 



(c) 



Figure 16.4 (a) nand symbol; (b) nor symbol; (c) not, or inversion, symbol. 



386 



Digital Circuits 



Note that the results of nand and nor operations are exactly the opposite 
of and and or, respectively. 

16.8 Logic Voltage Levels 

In order to explain the operation of actual logic-switching circuits, it is 
necessary to choose voltages to represent logic- 1 and logic-0 conditions in the 
gate circuits. Usually the choices arc made to give 

1. Positive logic, in which we make the 1 -logic level more positive than 
the 0-logic level. Frequently the assigned voltages are r-5 V for the I level 
and V or ground for the 0-logic level. 

2. Negative logic, in which we make the logic- 1 level more negative than 
logic 0, and voltages of —5 V and V are often assigned. 

Our descriptions of gale circuit operation will be based on the selection of 
positive logic. 

16.9 Diode Logic (DL) Gates 

Many different circuits have been designed to develop the gate functions 
described in Sec. 16.7. With diodes employed as switches, diode logic (DL) 
gates will first be discussed because of their simplicity. An and circuit is 
shown in Fig. 16.5(a); it is drawn with three inputs but the actual number 
may vary. 

+ 5V=K CC 



AND 
fl C Oulpul 




Output 










+ 5 V 
+ 5 V 






+ 5 V 
+ 5 V 







+ 5 V + 5 V 
+ 5 V +5 V 




+ 5 V 


+ 5 V 


+ 5 V 


+ 5 V 










+ 5 V 



Ao- 
tfo- 



Co- 



(b) 



ABC 



(O 
Figure 16.5 (a) Diode logic and gate; (b) ouiput table; (c) symbol. 



Diode Logic Gates 



387 



With positive logic voltage of I 5 V for the 1-logic level and ground for 
level, the diodes are connected so that 

1. Ifo// the inputs arc at logic 1 (4-5 V), the diodes have no voltage across 
them and are open ; the output voltage goes to I 5 V through ihe load resistor 
R. 

2. If any diode input is at logic (0 V), its diode is forward-biased by V cc 
and clamps the output voltage to zero or ground. This represents a logic-0 
output. 

These are the actions of an and circuit, as predicted by the table in Fig. 
16.5(b). The symbol for an and circuit, is presented in Fig. 16.5(c). 

Since the output voltage goes to -f 5 V through R and the diodes are then 
open, the output impedance is R at logic- 1 output. For logic-0 output, the 
output impedance is that of the sources supplying the conducting diode or 
diodes. The capacitance of the circuit load must be charged through these 
resistances and the rise time due to R will usually be longer than the time of 
fall of the output signal, when the discharge is through the low diode resis- 
tance. 

A diode logic or circuit is diagrammed in Fig. 16.6(a). For a positive logic 
signal the circuit will act so that 

1. if any diode input is at logic 1 ( I 5 V), that diode will be conducting 
and will place —5 V (logic 1) on the output. Two or more I 5-V inputs will 
produce the same effect as for one : 5-V input. 



.•1 o- 



OR 



«o M- 



C o- 



X 



Output 



(a) 



/; 










+ S V 
+ 5 V 
+ 5 V 
+ 5 V 






+ 5V 
+ 5 V 




+ 5 V 
+ 5 V 




+ 5 V 


+ 5 V 


+ 5 V 


+ 5 V 



(b) 



Output 





+ 5 V 
+ 5V 
+ 5 V 
+ 5 V 
+ 5V 
+ 5 V 
+ 5 V 



A o- 
I) o- 



C o- 




A +B + C 



(c) 

Figure 16.6 (a) Diode logic or gate; (b) output tabic; (c) symbol. 



388 Digital Circuits 

2. If all the inputs are at V or ground (level 0), all diodes are open and 
the output is connected to ground through R and the output is at logic 0. 

These are the actions of an or circuit, as shown by the table in Fig. 16.6(b). 
The circuit symbol for an or circuit is drawn in Fig. 16.6(c). 

In computing systems a circuit may be supplied by many inputs and must 
absorb the input currents. The number of inputs of similar circuits that can 
be connected without disturbance of the voltage levels is called the fan-in 
property of the circuit. The ability of a logic circuit to supply output current 
for driving other similar gates is indicated by the fan-out property or rating. 
A fan-out rating of 8 would indicate that a logic circuit could provide the 
current to drive eight circuits with similar input-current requirements. 

Due to the variable diode voltage drops, it is not possible to maintain the 
value of |5 V for the logic-1 level and this variability is a disadvantage of 
diode logic circuits. This variation is especially serious when many diode 
gates are used in series since the voltage losses become cumulative. The differ- 
ence between the 1 and levels may become insufficient for accurate operation 
of the circuits. Section 16.10 supplies a remedy for this situation. 

Reasoning as applied above will show that if negative logic signals are 
employed with these and and or circuits, the operating functions reverse; 
that is, a positive logic or circuit becomes a negative logic and circuit and a 
positive logic and circuit becomes a negative logic or circuit. Because of the 
ready availability of more satisfactory logic circuits, diode logic is now rarely 
employed. 

16. 10 The NOT or Inversion Operation 

The not function of inversion can be performed by a transistor in the C-E 
circuit in Fig. 16.7(a). The transistor is driven from saturation for an output 
at level to cutoff for an output at the logic-1 level, as shown by the output 
characteristic in Fig. 16.7(b). 

To ensure saturation, the base current should exceed the amount required 
at the intersection of the load line and the saturation line in Fig. 16.8; that is, 
the base current for saturation should be 

'rial i 



I„> 

The collector current at saturation will be 



(16.5) 



/r,..„ = '-££- 



V rr - V r 



R 



Vcc 
R 



since V CEi „„ < V cc . Then, to ensure saturation, we should have 



I B > 



h FE R 



(16.6) 









Integrated Logic Circuits 



f«-=+5V 




Oulpui V„ 



Output 




Vn < saI > 



- Logic I 



Output 







■ Logic 



(») 



I 
Logic Input 

(b) 

Figure 16.7 (a) Transistor inverter; (b) input-output curve. 




Figure 16.8 Saturation and cutoff levels. 

This will ensure a Iogic-0 output for a logic-1 input, as desired for the opera- 
tion of inversion. 

The output will swing from V C ei,,u = V for logic-0 output to V cc for 
logic-1 output. 



16.11 Integrated Logic Circuits 

Manufacturers produce many integrated circuit forms of the basic logic 
gates for and, or, not, nand, and nor operations. All will perform their 
stated functions and selection of a particular type depends on the applica- 






390 



Digital Circuits 



lion, with choice based upon the relative importance assigned to 

1. Speed of switching, usually stated in nanoseconds. 

2. Fan-in and fan-out characteristics. 

3. Power requirements and rated dissipation; usually in milliwatts. 

4. Cost, which usually increases with speed of operation. 

Each circuit is designed to perform one of the listed basic logic functions 
and an overall system represents the interconnection of many of the elemental 
circuits. Frequently four or six similar and independent circuits arc placed 
on one silicon chip and sold as quad or hex combinations to save space in the 

logic-circuit assembly. . 

" In general a manufacturer makes available a complete family or circuits, 
with compatible input and output voltage levels. The individual circuit forms 
arc simple, as will be shown. The internal circuit designs of typical units will 
be discussed here only to indicate the reasons for performance differences of 
the circuit types. 

16.12 Diode-Transistor Logic (DTL) 

In order to improve the fan-out capabilities of diode logic circuits and to 
maintain a standard logic- 1 level at the output, a transistor amplifier may be 
added to the output of the diode logic circuits and this leads to a circuit 
family known as diode-transistor logic (DTL). 

When the transistor is connected as an emitter follower, the output cur- 
rent is increased without loading the diodes. The transistor is more usually 
connected in the C-E circuit, however, and it then performs the not function. 
Since an and circuit followed by an inverter performs the nand operation 



Va 




Output 



Output 




(b) 



figure 16.9 (a) Diodc-lransistor nand circuit; (b) nor circuit. 



Diode-Transistor Logic 



391 



and an or circuit followed by an inverter yields a nor function, nand and 
nor circuits are available as illustrated in Fig. 16.9. These circuits with transis- 
tors are much less sensitive to output loading changes and have higher fan- 
out ratings than the diode logic circuits. Normal logic-level voltages are also 

restored at each gate. 

Resistors /?„ R 2 , and R } adjust the potentials between +K cc and -K cc 
so that in Fig. 16.9(a), when point * is at ground, the base of the transistor is 
placed well hclow cutolT and the circuit is immune to random noise pulses. 
With X at ground, current / is 

Vcc 



1 = 



rt 2 -i- Ri 



and 



Vbb = RJ - y cc 
Rx + Ri 



(16.7) 



'cc 



Thus when X is at ground, the base of the transistor is negative and the npn 
transistor is cut off. The output rises to +V CC or the logic- 1 level. When all 
inputs are at the 1 level, X is positive and the base of the transistor is driven 
to saturation. The output goes to V = 0.1 V, which is the logic-0 level, and 

the circuit is of NAND form. 

Another integrated circuit for DTL nand operation is shown in Fig. IfclU. 
Transistor 0. is driven by the diode output and supplies the saturation bias 




Figure 16.10 A DTL nand integrated circuit. 



r 



392 



Digital Circuits 



current for Q 2 ; g, is not driven to saturation. The speed is reasonably fast 
and the fan-out characteristics are good. 

A general disadvantage of DTL circuits is the necessity for two power 
supplies. 

16.13 Resistance-Transistor Logic (RTL) 

The resistance-transistor logic circuit family replaces the individual diodes of 
the DTL circuits with the base-emitter junctions of transistors; a nor circuit 
form is shown in Fig. 16. 1 1 as an example. The use of transistors instead of 
diodes does not increase the circuit complexity or cost when integrated circuit 
production methods are used. 



r cc 



:r, 



*3 

i — vww 



-M- 



€> 



R 2 

I — WW- 




Qi 



K 2 

i *Wvv- 




Output = 
A +fl + C 



Figure 16.11 RTL-nor circuit. 

Introduction of a sufficiently positive pulse (logic 1) to any input will 
drive the transistor to saturation. If two or more of the inputs are driven at 
the same time, the output voltage across the parallel transistors remains at 
the saturation level. The operation is that of or and this is inverted in the 
respective transistors. The result is that of a nor circuit, with logic-0 output 
for any logic- 1 input to A or B or C. 

Resistors R 2 are present to prevent the emitter-base junctions from short- 
circuiting the pulse sources. The input capacitance of the transistor must be 
charged through resistor R 2 and this introduces a delay. To allow fast pulses 
to charge the internal capacitance more rapidly, speedup capacitance C, is 
added to bypass R 2 for each transistor. Because of the time needed to dis- 
charge the transistor capacitances after saturation, however, the circuit is 
somewhat slower in operation than is the DTL form. The necessity for any 
one transistor to be able to lower the output from V cc to V CEI , U) means that 



Transistor-Transistor Logic 



393 



V cc must be small and the discrimination between logic- 1 and logic-0 levels 
at the output is limited. 



16. 14 Transistor-Transistor Logic (TTL) 

The transistor-transistor logic circuit family is illustrated by a TTL nand 
gate in Fig. 16. 12, adapted to integrated circuit manufacture. This circuit can 
be likened to the DTL circuit, except that the switching diodes of the DTL 
form arc replaced by separate emitters on the base electrode of a common 
transistor. If all the input emitters are given a logic- 1 pulse, Q t will be turned 
off and the input to Q 2 will rise, driving Q 2 into saturation with an output at 
V ceiuVI or the logic level of 0. 

Speed of switching is increased by use of the common-base form of circuit, 
increasing the high-frequency capabilities of the input transistor. The base of 
Q 2 is supplied through the low-resistance collector circuit of Q, and the 
stored charge in the input of Q 2 can be rapidly removed upon switching. The 
TTL family of circuits operates with switching speeds of 10 to 20 ns (nano- 
seconds); operating clock frequencies can be as high as 30 MHz. 

The gate output is either at logic 1, with Q, in saturation and Q, cut off, 
or at logic 0, with Q t cut off and Q y in saturation. Because of the manner in 
which the output circuit is usually drawn, it has been given the name of 
"totem pole" circuit. It is assumed that when Q 3 turns on, Q t turns off simul- 
taneously, and vice versa. Both transistors are on for part of the transition, 
however, and the output circuit then shunts current to ground. This current 
is limited by the low value of R. The result is that the circuit draws a spike of 



+ 5 V<?f tr 




F=ABC 



Figure 16.12 77X-NANr> circuit. 



394 



Digital Circuits 



current (a glitch) in each transition and this abrupt current drain may cause a 
spiked voltage drop on the V cc line, which is transmitted as electrical noise to 
other circuits supplied by that line. These noise pulses can cause random 
switching of circuits in which the voltage difference between logic 1 and logic 
is not great. If many gates switch simultaneously, a large peak current may 
be required from the power supply. 

In addition, the transition pulse of current represents power consumption 
and the power taken by TTL circuits increases with circuit operating rate, or 
the rate of the system clock. Under quiescent conditions the power may be 
only 6 mW per gate but at a clock rate of 20 MHz the power consumption 
may increase to 20 mW per gate. 

A great variety of related and compatible circuit functions is available 
for TTL system design. 

16.15 Emitter-Coupled Logic (ECL) 

Circuits 

When a transistor is in the saturation state, it has a base current higher than 
necessary and excess charge is stored in the base region and in the collector- 
base junction region. A transistor cannot change to the cutoff state until this 
charge is removed and the charge transfer slows transistor switching. Emitter- 
coupled logic (ECL) circuits obtain greater speed of switching by operation of 
the transistors in a nonsaturated condition, through limiting the lowest col- 
lector voltages to values above V CBUu) . An ECL circuit for the or, nor 
function is shown as an example in Fig. 16. 1 3. 




Figure 16.13 ECL-OR, NOR gate. 



CMOS Logic Circuits 395 

With fixed bias, transistor Q t supplies a constant current to the emitter 
resistor R E ; this current is less than the saturation collector current. A logic- 
1 input voltage to Q, will turn on that transistor and drop the output voltage 
at Z. As the switch transistor Q, increases its current, the increased bias 
across R E causes the current in Q 4 to fall and the drop across R E is maintained 
constant, but with current from Q, replacing the current from Q t . Reduced 
current in Q A raises the output voltage at Z and with outputs at Z and Z the 
circuit provides its own inverted output. 

The circuit is very fast, operating to switching speeds of 70 MHz; gives 
simultaneous or and nor outputs; and has high fan-in and fan-out capability. 
A complete logic family is available. 



16.16 CMOS Logic Circuits 

Because of major differences in the design and operation, the complementary- 
metal-oxide-semiconductor (CMOS) series of logic gates will be given more 
extensive treatment. 



+ K 



DD 



l\ S 



\U 



D 
D 



Q 2 Off j 



D 



<?i s 



V Dn = S V 



Q. On / Q. Off 



I Q 2 On 

\ 
\ 

\ 
\ 



X 



(a) 



2 
(b) 



Figure 16.14 (a) Basic CMOS inverter; (b) transfer characteristic. 

The CMOS switch employs FET enhancement-mode transistors in both 
/j-channel and w-channel forms, justifying the complementary name. The 
basic switch appears as an inverter in Fig. 16.14(a). This circuit shows/?- and 
/(-channel units connected in series across a power supply and with a com- 
mon gate input. These devices are diffused on a monolithic silicon chip, by 
processes that will be described in Sec. 16.18. 

In the ^-channel transistor Q 2 , the majority carriers are electrons. A posi- 
tive gate-to-source voltage greater than a threshold value V T will increase the 
channel current. For a gate voltage at ground or source potential, the channel 
is cut off. 



396 



Digital Circuits 



For the /^-channel unit Q„ the majority carriers are holes. A gate voltage 
negative to the source S increases the channel current; for the gate at source 
potential the channel is cut off. 

With the gates connected together, a positive input V„ equal to V Du , will 
turn on the //-channel Q 1 and turn off the /j-channel unit Q,. With the gates 
at ground, the //-channel gate is at source potential and less than V T ; Q 2 is 
cutoff; the gate of Q, is then negative to its source Sand <2, conducts heavily. 
The effect of a common input voltage to both gates is complementary and 
results in a push-pull action of the two transistor elements. 

Figure 16.14(b) shows the input-output transfer curve when the power 
supply is V DD = -f 5 V. When V, is near +5 V, the gate of Q 2 is at • 5 V to 
its source at ground and Q z is on. The gate-to-source voltage of Q, is then 
zero and less than the threshold voltage, however, so that Q, is in an off con- 
dition. The device is therefore operating in the right half of the transfer 
diagram; it appears to have a resistance of a few hundred ohms through Q 2 
to ground. The resistance of the off transistor will be in excess of 1000 MQ. 
The current through both devices in series will be less than 5 nA, resulting in 
very low power dissipation. 

We now place the gates near V or ground. The conduction conditions 
reverse, g 2 is off and Q, is on, and the output voltage reaches — V DD — • 5 V. 
Operation is in the left half of the transfer diagram. The quiescent current 
through both devices is again of 5-nA order. During the switching transition 
a small current passes as shown by the dashed curve on the transfer charac- 
teristic. 

With positive logic- 1 input, the output voltage is at V and in the logic-0 
condition; with Iogic-0 input, the output is at —5 V and in the logic-l con- 
dition. The device is therefore an inverter or not logic circuit. 

The switching time will approximate 25 to 200 nS. Because the circuit 
switches near one-half of V DD , its immunity to random circuit noise is good. 
Because of the very thin oxide layer that serves as gate insulation, the diode 
D is shunted between gate and ground to reduce negative overvoltages that 
might break down the oxide insulation. 



16.17 NAND and NOR Circuits with 

CMOS Logic 

By adding devices in parallel or series, the nor and nand circuits in Fig. 1 6. 1 5 
can be obtained with CMOS logic. In Fig. 16.15(a), if we make either input 
A or B positive, we turn on Q, or Q t ; but making A or B positive alS*o turns 
off either Q, or Q 2 and thus we have V„ = 0, or logic 0. We find the same 
result for both A and B positive. 

If we place A and B at ground potential or logic 0, transistors Q x and Qi 



NAND and NOR Circuits with CMOS Logic 



397 




+ 5 V 




(a) (b) 

Figure 16.15 Positive logic CMOS: (a) nor gate; (b) nand gale. 

are turned on and Q t and Q t are turned off; this is the condition for V = 
-f5 V, or logic-l level. 
The operation table is 



A 


B 


Vo 








1 





1 





1 








1 


1 






and we can identify this circuit as giving a nor output. 

The circuit in Fig. 16.15(b) is an upside down version of that in Fig. 

16.15(a). If A and B are positive or at logic 1, Q 3 and Q t arc turned on and 

V„ = 0, or logic 0. If cither A or B, or both, arc at ground or logic 0, then 

Q, or Q 2 arc on but Q 3 or Q 4 will be off and so the output becomes V = 

1 5 V, which is logic I. The operation table is 



A 


B 


v 








1 





I 


1 


1 





1 


1 


1 






ajxl the circuit is identified as giving nand performance. 



398 



Digital Circuits 



By adding an inverter at the output of either circuit we can develop or 
and and logic circuits. Additional inputs can be obtained by placing more 
transistors in series and parallel; the input capacitance increases and reduces 
switching speed. 

The power requirements are very low, being in the range of 20 /zW per 
gate for 15-V operation down to less than 1 /*W for 5-V operation. 

16.18 CMOS Manufacture 

The construction of a monolithic CMOS integrated circuit is illustrated by 
the cross section in Fig. 16.16(b). Manufacture starts with an n-silicon sub- 
strate wafer having a thickness of about 0.006 in. (0.015 mm). A silicon diox- 
ide layer is grown over the substrate by heating the silicon and a hole for the 
deep p diffusion at the left is etched through the oxide. Acceptor impurities 
are diffused into this large/? region to the desired depth. The original oxide 
layer is removed and a new layer grown overall. 

Holes arc etched open for the heavily doped p ■ diffusions at the right; 
these become the drain and source diffusions for the p-channel device. After 
the p~ regions are in place, the remaining oxide is removed and a new SiO : 
layer grown; holes are etched open for the heavily doped n !■ diffusions at the 
left. These areas serve as the drain and source electrodes for the /j-channel 
device. 

Oxide is again removed, regrown, and holes etched for the gate areas. 
A very thin oxide layer, about 10"" 4 mm thick, is grown over the gate regions 
and etched away from the metal contact areas elsewhere. Metalization is 
then added over the gate dielectric areas and for connection to the elements 
as shown. 



L 



nix 



D 
(a) 





n Channel — f" n Substrate T""'' ( ll '"" 1L ' 1 

Figure 16.16 Circuit and monolithic construction of the CMOS inverter. 



The Adder Circuit as a Logic System 



16.19 The Adder Circuit as a Logic System 



To show how these logic circuit elements are connected to produce a com 1 
puter subsystem, let us return to the addition process, analyzed in Sec. 16.6. 
If we identify logic-0 values by the complementary symbol X {X not), we 
can expand the logic statements of Eq. 16.1, showing that S = 1 when 

A and B and C, = 1 

OR 
A AND B AND C, = 1 

OR 

/and Band C, = 1 

OR 

A and B and C, = 1 

These statements can be made operative by use of inverters to obtain the 
complementary values, followed by four and circuits of three inputs each, 
connected to a four-input or circuit. The complete logic system for this 
portion of the addition operation is shown in the left portion of Fig. 16.17 
and yields the sum bit 5. 




Figure 16.17 Logical adder for one bit. 



400 



Digital Circuits 



The second statement in Eq. 16.2, for the forward carry C, = I, results 
in an expansion as 

A AND B AND C, = 1 

OR 
A AND B AND C, = 1 

OR 
A AND B AND C, = 1 

OR 
A AND B AND C, = 1 

These logic statements can be realized by three more and circuits feeding 
into a four-input or circuit. Since the last logic statement is identical to the 
last statement of the first set, the output of that and circuit is used and one 
and circuit is saved. The result appears in the right portion of Fig. 16.17 
and the circuit yields the carry bit C 2 . 

The complete circuit must be duplicated for each bit of the parallel- 
processed binary number, with appropriate interconnections for the carry 
bits C, and C 2 . Such complete adders are available on single integrated chips. 

16.20 Multivibrators 

The name multivibrator designates a group of circuits widely applied for 
switching as shift registers or temporary memories and as square-wave timing 
oscillators or clocks. The circuits are basically closed-loop feedback circuits 
operating with positive feedback. Because of the cumulative effects produced 
by this feedback, the circuits drive themselves to either of two limit or latch- 
up conditions, in which the transistors are at either cutoff or saturation. There 
are two output terminals A and B and when in a latched condition these 
terminals have potentials that are logic oppositcs as A = logic 1, B = logic 
0, for example. Switching can be made to occur between the two latched 
limits and the logic levels interchange so that A «= logic 0, B = logic I . 
Three general types of multivibrator are 

1. The bistable or flip-flop, which can be triggered from one stable latched 
condition to the other by an external signal. 

2. The monostablc multivibrator, which can be switched from one stable 
state to the other; it then returns to the first latch-up condition after a time 
delay. 

3. The astablc or free-running multivibrator, which continuously switches 
between its two limits without application of an external signal; it is a square- 
wave oscillator. » 



The Bistable Multivibrator or Flip-Flop 



401 



The flip-flop is of great importance in digital operations; the monostablc 
is useful as a delay and timing circuit; and the free-running type is used for 
the timing oscillator or clock of a computer system, as well as a frequency 
source. 



16.21 The Bistable Multivibrator 

or Flip-Flop 

The circuit in Fig. 16.18(a) illustrates the operating principle of the important 
bistable multivibrator or flip-flop. The circuit employs two inverter amplifiers 
in a positive feedback loop, with the output of inverter 1 fed to the input of 
inverter 2 and the output of inverter 2 returned to augment the input of 
inverter 1. The two stable or latched conditions occur with inverter 1 in 
saturation and inverter 2 in cutoff and the reverse with inverter 2 in saturation 
and inverter I at cutoff. The outputs are complementary and if point A is 
high at logic 1, then B is low at logic-level 0. If an external triggering signal is 
applied, the circuit may be made to switch and latch in its reverse state with 
A at low potential and logic-level and B high at logic-level 1. 

For further understanding, a form of discrete circuit is drawn in Fig. 
16.18(b). Assume the circuit is in a latched state with Q l at cutoff and Q z in 
saturation. The voltage at A is equal to V cc , while that at B approximates 
zero. Cross coupling from the low voltage at B through R z and R, and aided 





B s 



(a) <b) 

Figure 16.18 (a) Bistable inverter loop; (b) circuit of bistable multivibrator. 






402 



Digital Circuits 



by the bias — V cc maintains the base of Q x at cutoff voltage. Cross coupling 
from A through R, and R, places a positive potential on the base of Q 2 . 
Thus Q, is held at cutoff and Q 2 in saturation and the circuit is actually in a 
stable or latched state. 

A negative trigger pulse may then be applied to R, the reset terminal, and 
to the base of Q 2 , driving that transistor toward cutoff. The voltage at B 
rises as the Q 2 collector current drops and by coupling through R 2 the poten- 
tial rise at B is transmitted to the base of Q„ turning that transistor on. The 
rise in its collector current then reduces the voltage at A ; this fall of voltage is 
transmitted through R, to the base of Q 2 and drives that transistor further 
toward cutoff. This action raises the voltage at B further and the cumulative 
action proceeds around the loop until the second latched state is reached, 
with Q, in saturation and Q 2 at cutoff. The voltage at A is then near zero or 
at logic and the voltage at B is V cc and at logic 1. A negative pulse applied 
to S, the set terminal, will cause the circuit to return to its original state, with 
A at logic I and B at logic 0. 

The time required to transfer conduction from one state to the other is 
known as the switching time. Fast switching is possible if the transistors are 
chosen with high values of/V, indicating low values of input capacitance. The 
time constants associated with the speedup capacitors C should be small with 
respect to the time between switching pulses to allow the speedup capacitors 
to recharge. 

16.22 The RS Flip-Flop 

Figure 16.18(b) shows the R and S trigger terminals of a reset-set flip-flop. 
In Fig. 16.19(a) we have a circuit composed of a cross-coupled pair of nand 
gates for the multivibrator switch, with input logic supplied by two additional 
nand gates. This circuit will act as does an RS flip-flop, although the actual 
circuit may be much more complex. 




s 




Q 


r 


IT 




R 




Q 



(a) » (b) 

Figure 16.19 (a)NAND gale simulation of an RS flip-flop; (b)/?S flip-flop symbol. 



The T or Toggling Flip-Flop 



403 



The circuit will trigger on the positive-going edge of the clock pulse in- 
serted at Cif a positive I exists at the S or R inputs. A logic- 1 input to 5 with 
the input to R at will cause Q to go to I and Q to 0; this is the set state. If R 
is given a logic- 1 signal when S is at 0, the output condition becomes Q = 
and (2=1; this is known as the reset state. 

Simultaneous inputs to S and R will not give outputs from the nand 
logic circuits when a clock pulse arrives and the flip-flop is left in its previous 
condition. Simultaneous 1 inputs to R and S will cause circuit confusion, 
however, since the circuit will attempt to reach logic on both outputs. This 
is an input condition that must be avoided. 

The circuit operations arc predicted by 









s 


R 


Q Q 








Unchanged 





1 


1 


1 





I 


1 


1 


? ? 



In effect the RS flip-flop indicates and remembers the last input signal 
received. The switching action is synchronized with the arrival of a positive- 
going clock pulse. Input pulses need be only long enough to initiate the 
switching action programmed by the R or S signal and the input pulse can 
then be removed. 



16.23 The T or Toggling Flip-Flop 

The T multivibrator in Fig. 16.20 will switch between alternate states on 
successive negative trigger pulses. The two diodes D x and D 2 arc steering 
diodes, which direct the trigger pulses to the base of whichever transistor is 
on. A reset input may also be added to the base of a transistor to set the initial 
condition of the circuit before toggling starts. 

With Q, off and Q 2 on, a - 5-V negative trigger pulse is applied and takes 
point X down to f 5 V. Diode D 2 is reverse-biased between C at -| 5 V and 
Bat V. Diode D, is found to be forward-biased and transmits the negative 
trigger pulse past A to the base of Q 2 , turning it oflT. Transistor Q t is turned 
on in normal fashion. 

The next negative pulse finds D, reverse-biased due to the change in 
voltages at A and B and, with D 2 conducting, the trigger pulse is directed to 
the base of Q, to turn it off. The circuit then toggles or reverses its outputs. 

Toggle flip-flops are used in counters to accumulate pulse counts. Each 
toggle circuit is known as a scale-of-two circuit since Q 2 switches on once for 



404 



Digital Circuits 



Trigger 



o!' rr = + 10 V 






(b) 






la) 



Figure 16.20 The logglc flip-flop. 



Iwo input pulses. In switching on, Q z generates a negative-going pulse at B 
that can be transmitted to a second toggle circuit as a triggering pulse. For 
two toggle circuits in cascade a pulse is obtained from the last circuit for every 
2 2 = 4 input pulses. This is the action of a scale-of-four circuit, as shown in 



Input 
o If- 



C\ 



TT 



.v cc 



C 





*2 

Y 






/>4 



-o Output 



/) 



Figure*16.2I Scale-of-four circuit. 



The JK Flip-Flop 



405 



Fig. 16.21. The count can be extended by use of additional toggles in cascade, 
n circuits providing for a count of 2". 

If positive logic signals are used, then the negative-going input triggers are 
obtained when the trigger pulse goes from -|-V to V and this is known as 
trailing-edge triggering. 

16.24 The JK Flip-Flop 

The JK flip-flop is the most complex of these circuits. It is similar to the RS 
circuit except that its operation is predictable when both trigger inputs arc at 
the logic- 1 level. In that condition the circuit always changes state and com- 
plements its previous outputs. 

A circuit that will provide JK operation is drawn in Fig. 16.22; actual 
circuits arc more complicated but being available in integrated circuit pack- 
ages makes this internal complexity of little importance to the user. 

Since each and gate has one input fed back from an output, one of the 
and gates has a logic- 1 input in addition to the logic I supplied by the clock 
at T. We also have the J and K inputs to the respective gates and four possi- 
bilities exist; these are listed in the operation table: 



J 


a: 


Q Q 








No change 





I 


1 


1 

1 




i 


I 
Toggles 



Signals may be applied and held at J and K but switching occurs only at 
the time of the positive-going rise of the clock pulse applied at T. 




J 




a 


1 


FP 




K 




Q 



(a) (b) 

Figure 16.22 (a) Equivalent circuit for a JK flip-flop; (b) symbol for a JK 
circuit. 



406 



Digital Circuits 



As seen from the table, if both J and K are at logic the and gates arc 
inoperative and no action occurs when the clock pulse arrives. If both J and 
K inputs are at logic 1, the action of the clock pulse is as a toggle and the 
output becomes the complement of the preceding condition. The circuit then 
operates as a T flip-flop. 

For./ = 1, K ^- or/ = 0, K = 1, the outputs follow the inputs and the 
circuit operates as an RS flip-flop. 

Figure .16.23 illustrates a JK flip-flop as designed with CMOS switches. 
The transistors designated Q, and Q 2 constitute the basic nor circuit of the 
RS form of flip-flop, with Q 3 , Q 4 providing the JK features and Q, supplying 
the clock toggle action. The complexity is not a matter that need concern us 
when the circuit is designed as an integrated package. 



1 



DD 



»h 



Q 



<?o- 






«»Z3\- 






H 



fl»» 



^ 



Jo- 



*r ^ 



h 



-oQ 




-oK 



-or 



Figure f 16.23 A CMOS JK flip-flop. 



The One-Shot or Monostable Multivibrator 



407 



16.25 The One-Shot or Monostable 

Multivibrator 

The basic circuit of the monostable or one-shot multivibrator is derived from 
the bistable form of circuit by replacement of one of the cross-coupling 
resistors with a capacitor C. The result is indicated in Fig. 16.24 where we 
have two inverters in a closed loop, with the positive feedback path between 
the output of inverter 2 and the input to inverter 1 consisting of the ac path 
provided by capacitor C. 



In o. 




on 




^( o Trigger 



(a) CM 

Figure 16.24 (a) Cross-coupicd inverters acting as a monostable multivibrator; 
(b) a monostable circuit. 

The normal stable condition finds inverter 1 at cutoff. If a positive pulse is 
given to inverter 1, driving it into saturation, its inverted output will drive 
inverter 2 to cutoff. The signal is coupled back to the input of 1 and will hold 
inverter 1 in saturation as long as the charge remains on C. This is a quasi- 
stable state for the circuit and after the charge drains ofl" C, the circuit switches 
back to its original stable state in which inverter 1 is cut off. 

The time duration Tin which the circuit is in the quasi-stable condition is 
determined by C and its associated resistances and the delay time in that state 
can be varied by means of a variable resistance. The one-shot circuit is fre- 
quently used to provide a delayed pulse or to vary the length of a received 
pulse. It is then known as a "pulse stretcher." 

A simplified circuit for a monostable multivibrator is drawn in Fig. 
16.24(b), with one cross-coupling resistance replaced with capacitor C as the 



408 



Digital Circuits 



feedback element. Normally Q 2 is in saturation by reason of the positive bias 
supplied through R and Q, is cut off because the voltage at fiand the base of 
Q x is essentially zero. 

A negative trigger pulse will turn Q 2 to the off condition. As its collector 
current drops, the voltage at B and the base of £>, rises to V cc , turning Q, 
on. As the Q, current rises, the voltage at A falls from I V cc to zero and this 
negative step potential is transmitted through C to the base of Q u placing it 
in the off condition. This change in base voltage is sketched in Fig. 16.25(b). 

Transistor Q z is held off until capacitor Ccan recharge and the time delay 
is indicated as T seconds. When the base voltage of Q z rises to zero, Q 2 
turns on and Q, goes to cutoff and the original stable condition is restored. 

The time of delay is determined from the RC time constant as 

T^OJRC (s) (16.8) 

The outputs at A and B are complementary. 



Trigger 
Pulse 



n 



t = 



(a) 



To V. 




a 



T Delay 



II +V 



tfiBi) 1 



(b) 



<c) 



Id) 



Figure 16.25 Circuit waveforms for a monostablc multivibrator. 



16.26 The Astable Multivibrator 

The third version of the multivibrator circuit is the astable or free-running 
form, which acts as a square-wave oscillator. The circuit is used to provide 
the square-wave clock signals for digital processing. 

The astable circuit is derived from the bistable form by replacing both 
feedback resistors with capacitors, as indicated in Fig. 16.26(a). With feed- 
back supplied by capacitors, there is no stable state for the circuit and it 
oscillates with a square-wave output at B and a complemented square-wave 
output at A, as shown in Fig. 16.27. 















Vcc 

.1. 
















:*/. «i 




r 2 k, : 






1 




o,-i 


A' 


K 




If 


II 






c : 


\ 


X 


•*"c, 




Q,( 








JQi 






j*j ■ 


° II 













(a) (b) 

Figure 16.26 (a) Astable action by feedback with inverters; (b) stable multivi- 
brator circuit. 




+ Vcc 



/ = 



"Bt: 



+ K 




(a) 



(b) 



(O 



° r=~0 (d) 

Figure 16.27 Waveforms of the astable multivibrator. 



409 



410 



Digital Circuits 



Consider Q t just turned off at / — in Fig. 16.27(a) and (b). The base 
voltage of Q, will rise from — V^ toward zero as C, charges. When v BB , 
reaches V, Q, turns on and turns off Q 2 , as shown in Fig. 16.27(c) and (d) 
at / = r,. The cycle repeats as C 2 charges, after which Q, and Q z again 
switch, returning to the original condition. 

The oscillation continues and the frequency is determined by the time 
constants R X C, and R 2 C 2 . These determine the respective time delays and 
since these need not be identical, the frequency of the oscillation can be 
stated as 

(16.9) 



/ = 



r, + r, o.7(*,c, -i r 2 c,) 



(Hz) 



75.27 Synchronization of the 
Free-Running Multivibrator 

In the freely oscillating form, the rectangular output waves of the multivibra- 
tor circuit include many harmonic frequencies. This was the reason for the 
multivibrator name for these circuits; however, the stability of the frequency 
generated is not great. The circuit can be synchronized lo a standard frequency 
or pulse chain by applying pulses to the base of the off transistor so that these 
pulses add to the rising base-voltage wave, as in Fig. 16.28. The added pulses 
cause triggering of the circuit at a time determined by the pulses rather than 
by the rising charging curve. 

The synchronization need not be at the pulse rate but may take the form 
of frequency division as shown, in which the multivibrator triggers for every 
n pulses. The figure shows a countdown ratio of 8. The harmonics in the 
square output waves then provide standard and accurate frequencies covering 
a wide frequency range. 

The horizontal- and vertical-sweep oscillators of TV receivers are syn- 
chronized in this manner to pulses received with the picture signals. 

I I I I I I I I I I I I I 



<:.) 



Vae 



K V 



(b) 



Figure 16.28 (a) Synchronizing pulses; (b) countdown by a factor of 8. 



The Schmitt Trigger Circuit 



411 



16.28 The Schmitt Trigger Circuit 

The Schmitt trigger shown in Fig. 1 6.29 replaces one of the cross-coupled 
feedback paths of the multivibrator with feedback coupling across a common- 
emitter resistor R E . The circuit switches at two input voltage levels, one 
when the input voltage is rising toward the trigger level and the other when 
the input voltage is falling from above the trigger level. The difference in 
these levels is called the hysteresis of the circuit. 

The triggering voltage level is set by V E , the voltage across the emitter 
resistor R E and due to the current of Q 2 . For input voltages less than V E the 
input transistor is cut off by this emitter bias, Q 2 being in conduction with a 
low voltage at B. As v, approaches V E , the collector current of Q, starts to 
rise, lowering the voltage at A as well as the base voltage of Q 2 . As Q, 
turns further on, Q 2 cuts off because of the feedback through /?, from A and 
the voltage at B rises to V cc . A reverse action follows when the input voltage 
falls from a value above the original V B . 

The change of output voltage at B versus the input v, is plotted in Fig. 
16.29(b) and shows the hysteresis in switching voltage that exists when the 
input voltage falls from above the V E value. By always approaching 
the switching level from below V E , the operation will accurately occur at 
the same input voltage. 

Thus a switched voltage is available at B when v, = V E , and this voltage 
change can be used to indicate equality of the voltages for voltage comparison 




£ 


Vcc 






3 




1 


\ 


Q. 




1 




3 




1 




o 




1 
1 
1 

t 


1 
1 

1 

1 







Input V, 



(a) (b) 

Figure 16.29 (a) The Schmitt trigger; (b) input-output relations. 



412 



Digital Circuits 



Decoding Matrices 



413 



circuits. The circuit also will provide squared output waves for sinusoidal or 
other pulse forms. In particular, it is used to reshape pulses that have been 
distorted in transmission, as shown in Fig. 16.30. 



Input V 



; -/A--/A-/A- 



(a) 




(b) 



.2J-L- 

Figure 16.30 Reshaping of pulses by (he Schmilt trigger. 

16.29 The Shift Register 

One of the most common applications of the flip-flop is in the shift register. 
This is a form of temporary memory in which data pulses, I and levels, can 
be transferred serially from one flip-flop to the next flip-flop. Readout of the 
condition of all flip-flops can be made on call, with the data flowing into 
parallel channels for further processing in parallel form. 

In the circuit of Fig. 16.31(a), serial data is continuously supplied from a 
telephone line or other data channel into the first flip-flop FF,. The input 
pulse is complemented by an inverter and the pulse and its complement are 
fed to two and circuits along with the clock pulses. A I input pulse present 
at the time of the first clock pulse goes into the and circuit along with the 
clock pulse and the upper and circuit output feeds a logic I to the set input of 
FF,. If a logic had come from the line, it would have been inverted and sup- 
plied as a 1 to the lower and circuit along with the clock pulse and a I output 
would then have gone to the R input, resetting FF, with a at Q. 

On the next clock pulse the conditions at the outputs, Q and Q, will be 
transferred through the second set of and circuits to S and R of FF Z ; new 
data will be supplied to the inputs of FF,. On the third clock pulse the con- 
dition of FF 2 is transferred to FF t , the state of FF, is transferred to FF ly and 
the new input pulse enters FF,. The input data pattern flows through the 
register with identical waveforms, as shown in Fig. 16.31(b). 

The first stage samples the input data at the instant of the trailing edge of 
a clock pulse; even though the data input changes state at random times, the 
data at the output of each flip-flop changes only in synchronism with the 
clock. 

If four-bit numbers are used, then gates placed at the outputs 1, 2, 3, and 
4 can be activated once every four clock pulses and the bit values of the four- 
bit number then present in the register will be read into four parallel chan- 
nels, for translation of serial input data for parallel processing and further 






Input 

Datu 

o 







Clock 
Pulse 



(a) 



I 2345678° 10 

«- JUUUUUUUULJl 



Input 
Data 




Cb) 

Figure 16.31 (a) A shift register; (b) flow of signals through the register. 

computation. The addition of more flip-flops to the register would increase 
the number of bit positions that could be handled from the input data. 

The shift register can also be used as a delay element. Input pulses may be 
fed serially to the input of a register of appropriate length and the same 
data taken out of a later flip-flop, delayed by the time of one clock pulse per 
stage included in the register. 

16.30 Decoding Matrices 

A decoder matrix is usually a diode logic network, actuated by a shift register 
as a temporary storage clement, and used to decode binary signals to decimal 
or other code. The application in Fig. 1 6.32 is intended to translate three-bit 
BCD numbers into decimal outputs 0, . . . , 7; another flip-flop in the register 
and more diode logic switches would allow extension to decimal 15. 



414 



Digital Circuits 



Decimal Counting 



415 



+ V 



Binary 

in Pww 



no -ww 



IOI -WW 



IOO -WW 



on -ww 



010 -WW 



001 -WW 



ooo 




Shift 
Register 



Serial 
Input 



Figure 16.32 Decoding matrix of and circuits. 



The outputs of the flip-flops in the shift register are applied to lines con- 
necting to diodes, each vertical bus constituting a four-input positive logic 
and circuit. With -| V connected to the horizontal lines, any diode connected 
to a Q — flip-flop state will conduct and ground that particular horizontal 
bus. For any state of the flip-flops, only one bus will not be at zero potential 
and that bus will place — V at the terminal corresponding to the decimal 
equivalent of the binary number to be indicated. 

A similar diode matrix can be designed for any input code by noting that 
a 1 in the binary number calls for a diode at a Q connection ; a in the binary 
code requires that a diode be connected at Q. 

16.31 Decimal Counting 

A chain of four flip-flops will count to 2* ■= 1 6 pulses before the last flip-flop 
returns to its initial state. When using decimal numbers, it is more convenient 
if we can count by decades and modifications of the basic scaler circuits are 
available, using feedback, to cause them to count by 10's. 







Feedback on 
-» I Change 



Figure 16.33 A divide-by-10 feedback counter. 

One such circuit is shown in Fig. 16.33. Initially a reset pulse is applied to 
place all the flip-flops in the normal condition. Counting is normal up to the 
eighth pulse, and the conditions of the flip-flops are indicated in Table 16.3. 
At the eighth pulse, Q of FF, goes from to I and this transition provides a 
pulse that is fed back to FF Z and FF } to advance them by one pulse each. 
Including the weight of 2 for FF t and 4 for FF„ the counter is therefore 
advanced by a total count of 2 + 4 — 6. With the count of 8 already present 
at the output of FF t and an artificial advance of 6, the count now appears to 
be 8 + 6 = 14; two more input pulses advance the output to 16, giving a 



TABLE 16.3 



Pulse No. 


(?. 


Q\ 


Qz 


Qi 


Qi 


G, 


Q* 


Q* 





1 





1 





1 










1 





1 


1 





1 










2 


1 








1 


1 










3 





1 





1 


1 










-4 


1 





J 








1 







5 





1 


1 








1 







6 


1 








1 





1 







7 





1 





1 





1 







8 


1 





1 





1 








1 


8* 


1 








1 





1 





1 


9 





1 





1 





1 





1 


10 


1 





1 





1 





1 






•By feedback. 



416 



Digital Circuits 



to l transition at Q of FF t , an output pulse, and resetting all flip-flops to the 
initial zero count condition. 

Actually the output pulse is transmitted after 

N » 2« - 6 = 10 pulses 

and so we have a divide-by- 10 counter. Additional circuits can be added in 
cascade for counting of further decades. 

The condition of the individual flip-flops is indicated in Table 16.3. 

At the tenth pulse the flip-flop conditions correspond to a zero count and 
a to I transition pulse is transmitted from Q t . 

Such decade counters are available as complete integrated circuits. Many 
variations of the basic scheme are employed as elements in frequency coun- 
ters. 

REVIEW QUESTIONS 

16.1 Explain the meaning of zero in a decimal number. 

16.2 Why is a binary code well adapted for use with transistors? 

16.3 What is the weight assigned to each position in the binary number 0101 ? 
What is the decimal value of that number? 

16.4 Convert decimal 812 to its binary equivalent. 

16.5 Convert decimal 0.764 to its binary equivalent. 

16.6 What is the meaning given to the word bill 

16.7 In decimal numbers, what is the 10's complement of 371 ? 

16.8 In binary numbers, what is the 2's complement of 1001 10? 

16.9 Write the BCD number for decimal 1028. 

16.10 What is the decimal equivalent of the BCD number written as 

0111 0101 1000 

16.11 How many bits arc needed to transmit the decimal number 137 as a binary 
number? 

16.12 What is a byte? 

16.13 Write decimal 317 in octal code. 

16.14 What is meant by parallel processing of digital signals? 

16.15 A signal contains 32 bits per byte. If the time to process one bit is 1.1 /IS, how 
long does it take to process one byte in series and in parallel operation? 

16.16 Describe the operation of a two-input and circuit. 

16.17 Describe the operation of a two-input or circuit. 

16.18 In logic algebra, what is the meaning of the symbol • ? Of the symbol +? 

16.19 What is a nand circuit? A nor circuit? 

16.20 Define positive logic; negative logic. 



Problems 



417 



16.21 Two voltage levels of -f-5 V and — 5V are available in some equipment. 
How would you assign these in 

(a) Positive logic? 

(b) Negative logic? 

16.22 What is meant by the fan-in of a circuit? 

16.23 What is the meaning of fan-out? 

16.24 What is a logic gate? 

16.25 Name one disadvantage of DL circuits. 

16.26 What happens to and and or circuits when the logic voltages are reversed? 

16.27 What is a not operation? 

16.28 Name at least four factors to be considered in the selection of a particular 
form of logic circuit. 

16.29 Name one advantage of DTL over DL. 

16.30 Name one advantage of TTL circuits over DTL; also name one disadvantage. 

16.31 Name one operating advantage of the ECL form of circuit. 

16.32 To what design factor does the ECL family owe its switching speed? 

16.33 Describe a CMOS inverter switch. 

16.34 Why is the power consumption of a CMOS switch so small? 

16.35 What is meant by a flip-flop? 

16.36 What is a monostable multivibrator? 

16.37 What is an astable multivibrator? 

16.38 What is meant by multivibrator-switching lime? 

16.39 Why arc speedup capacitors used? 

16.40 What are the output conditions represented by a trigger pulse of a flip-flop 
to the set terminal ? To the reset terminal ? 

16.41 Describe the action of an RS flip-flop. 

16.42 How docs a JK flip-flop differ from an RS flip-flop? 

16.43 What is the function of a T flip-flop? 

16.44 What is the cause of the quasi-stable slate in the one-shol multivibrator? 

16.45 What is the purpose of synchronizing a free-running multivibrator to a stan- 
dard-frequency oscillator? 

16.46 How would you use a Schmitt trigger? 

16.47 What is the purpose of a shift register? 

PROBLEMS 



16.1 Decode the following binary numbers to decimal form: 

10110 01010 001 1 1 
11010 10101 01011 

oino 11000 nioi 



418 



Digital Circuits 



Problems 



419 



16.2 Write the following decimal numbers as binary numbers: 

104, 547, 123, 362, 445, 176 

16.3 Translate the following decimal numbers to base-8, or octal, numbers: 

123, 387, 462, 97 

16.4 Translate the following decimal numbers to base-3 numbers: 

57, 96, 81, 104 

16.5 Perform the following operations in binary arithmetic, with -I meaning 
addition and x implying multiplication: 

110110 I 01 101 1 (a); 1 10010 x 10110 = (c) 

101 111 | 100101 - (b); 101010 x 11100 =(</) 

16.6 The following numbers are in binary notation: 

01011101 01101100 0.1101 
10110010 11100011 0.0111 
1 101 101 1 10(01 100 0.101 1 

(a) Determine the decimal equivalent values. 

(b) Determine the equivalents in octal notation. 

16.7 For the indicated operations in Fig. 16.34(a), write the output expression in 
words as F A or B 











D>- 



(a) 



(b) 



/lo- 



Ho ■£ 




/lo- 



C"o- 






(c) 



Figure 16.34 



16.8 Implement the operations of Fig. 16.34(a) with diode-transistor logic circuits 
and positive logic, and draw the circuit. 

16.9 Derive the operation table for the circuit in Fig. 16.34(a) by writing the table 
step-by-step. 

16.10 Repeat Problem 16.7 for the circuit in Fig. 16.34(b). 

16.1 1 Draw the circuit, using DTL gates, to carry out the operation in Fig. 16.34(b). 

16.12 For A = 0, B = 1, find the output state F of the circuit in Fig. 16.34(b). 

16.13 The waveforms in Fig. 16.35(a) are applied to a nor gate. Draw the output 
waveform. 

16.14 The waveforms in Fig. 16.35(a) arc applied to an and gate. Draw the output 
waveform. 



+ 6V 



+ 5 



•JTTLTL 



+ 5 



"2 



J 



r 




(a) 



(b) 



Figure 16.35 



16.15 Both inputs to the RTL nor gate in Fig. 16.35(b) are at +6 V. If R, = R z 
= 10,000 fi, R L -■ 1000 CI. what arc the collector currents of Q, and Q 2 
if h fB - 50? Consider the R L drop. 

16.16 In the circuit in Fig. 16.35(b), each transistor has a reverse saturation current 
of 20 fiA in the off state. How many similar inputs could be added before 
the off-state output voltage drops to 5.5 V? Use the circuit resistances of 
the previous problem. 

16.17 Consider the nor gate in Fig. 16.35(b). How much current can be supplied 
to a load at »>„ if v„ is not to be less than 5 V for logic I, the gate transistors 
being in the off state. 

16.18 For the RTL nor gate of Problem 16.17, if v, = +5 V, what is the input cur- 
rent? Will that input current saturate Q, if h rE = 30, V cc = 5 V? 

16.19 Write the operation table for the circuit in Fig. 16.34(c). What is the output 
state when A I, B - 0, C = I ? 

16.20 Determine if the two circuits in Fig. 16.36 are equivalent in operation by 
comparing their operation tables. 



420 



Digital Circuits 



Problems 



421 



/lc 
flc 

/Tc 
flc 




/lc 
Be 

/Tc 

flc 



+ 4 V 






(a) 



(b) 



Figure 16.36 



16.21 Figure 16.37(a) shows a CMOS bistable multivibrator. What are the stable 
operating conditions for the transistors of this circuit? 

16.22 (a) Explain the operation of the circuit in Fig. 16.37(b). 

(b) What is the purpose of Q x and gj? Note that the circuit does not employ 
complementary symmetry. 



0^ 




r DD 



X 



e 2 



K 



3 




C>3 



R 



(b) 



Figure 16.37 



16.23 Determine the output frequency from an astable multivibrator having timing 
circuit component values of /?, = 2.7 kQ, C, 1000 pF and R 2 =5.2 
kQ, C 2 - 1500 pF. 

16.24 The RTL bistable circuit in Fig. 1 6.38(a) is to be changed to monostable form. 
Draw the circuit and give possible values for R and C to give an off time of 
10 /xs for Q 2 . 

16.25 Show whether the inverter in Fig. 16.38(b) is driven into saturation If Am = 
20, with logic levels of +6 and V applied at A. 







<-6 v 




o l'„ 



(a) 



(b) 



Figure 16.38 



16.26 If a second trigger pulse is applied to a one-shot multivibrator while it is in 
the quasi-stable state, what will be the elTect on operation? 

16.27 The excess-3 code is sometimes used because no digit is represented by a com- 
plete null or zero signal : 






0011 


5 


1000 


1 


0100 


6 


1001 


2 


0101 


7 


1010 


3 


0110 


8 


1011 


4 


01 1 1 


9 


1100 



Design a diode-switching matrix to translate exccss-3 coded numbers to their 
equivalent decimal values. 



r 



17 

Power Control 



The basic design of a junction transistor is not well suited for switching in 
high-power applications. For efficiency with high currents, the emitter and 
collector regions should be of low resistance or of low resistivity material; 
for high h rB , the base should be very thin. A thin base between a low-resis- 
tance emitter and collector is not suited to operation at the high voltages 
needed in industrial switching. By use of four material layers and three junc- 
tions in series, however, the silicon-control rectifier (SCR) family of devices 
has been developed to control high currents at high voltages. This family 
includes various related designs known as the SCR, thyristor, triac, and 
the gate-controlled switch. 

17. 1 The Silicon-Control Rectifier (SCR) 

In contrast to the continuous control of current in the transistor, the silicon- 
control rectifier action is that of a trigger able to switch a current on. The 
current can only be stopped by reducing its magnitude below a certain hold- 
ing value /„. The four-layer pnpn construction of an SCR is shown in Fig. 
17.1(a). 

We can consider the unit to consist of two transistors in series, as dia- 
grammed in Fig. 17.1(b). Transistor Q, is composed of layers n t ,p l ,n l and 
is of npn characteristics and transistor Q, employs layers Pz,n t ,p x and is of 
pnp characteristics. A gate electrode is connected to/>, . With the gate grounded 



422 



SCR Characteristics 



423 



or V a = 0, transistor g, is cut off and its collector current I Cl = I co and 
consists of the leakage current only. This is the base current to Q z and it is 
too small to turn that transistor on. The device is open-circuited for V = 0. 

When we apply a sufficiently large positive voltage to the gate, we turn 
Q, on. The collector current is l Cl — I Bl and so Q 2 turns on; but the collector 
current of Q z is I c , — I B , and so Q, turns on further, increasing I Cl = I Bl . 
The actidn around the Q u Q z loop is cumulative and the device rapidly 
switches to a conducting condition between anode and cathode. With both 
bases driven to saturation the forward resistance between anode and cathode 
is very small. The time required for the cumulative turn-on action is ap- 
proximately 0.1 to 1 p.s. 

Except for the initial triggering action, device conduction does not depend 
on the gate current. Therefore turn-off does not occur when the gate signal 
is removed since the two base currents are now internally driven. To stop 
conduction the anode voltage must be removed or the anode driven to a 
negative potential, reducing the current below /„. Turn-off time is typically 
5 to 30 fis. 





9+V 

An 




Pi 




"2 


Gate 


P\ 






"I 



I 



Cathode 




Cathode 



(a) (b) 

Figure 17.1 (a) A four-layer switch; (b) two-transistor simulation of four-layer 
action. 

77.2 SCR Characteristics 

The SCR is normally used with an ac voltage applied and conduction occurs 
on the half cycle in which the anode is positive. The characteristic for one 
value of gate current 1 is shown in Fig. 17.2(a), with the circuit symbol of 
the SCR presented in Fig. 17.2(b). 



424 



Power Control 





/ 


C < 


/ 


- V 


b 


- Z 1 

/ 


:x^ B 


f 







V, + V 


1 Zener 








1 Region 









9 Anode 



J: 



Gate 




< Cathode 



(a) (b) 

Figure 17.2 (a) Voitagc-currcnt relation for the SCR; (b) symbol. 

For a particular value of gate current the SCR will trigger when the volt- 
age of the wave reaches V B . For larger values of /„ the triggering voltage will 
be lower. As conduction starts at point B on the voltage-current curve, the 
SCR voltage abruptly falls to Cand then rises to its peak value along the CD 
portion of the curve. In the on condition the SCR acts as a forward-biased 
diode and the curve between C and D is that of a junction diode. A 
great many combinations of gate voltage and gate current will trigger the 
SCR but gate voltages greater then 3 V arc usually required. 

After the peak of the ac current wave is passed on the CD curve, the 
current value falls along the curve; when it reaches I h , the holding current, 
the conduction ceases and the current goes to zero. With ac applied, con- 
duction ceases when the ac sine wave approaches zero. 

Applications with ac involve the use of a pulse of gate current /„ at the 
time it is desired that conduction start in the circuit in Fig. 17.3(a). This will 
beat some angle #, in the wave of ac anode voltage illustrated in Fig. 17.3(b). 
The average or dc value of the rectified current varies with the triggering angle 
6, as shown in Fig. 17.4. For a half-wave-controlled rectifier circuit with a 



AC 



J'. 



Gate 



v. i 




SCR 




o VvW 

(a) (b) 

Figure 17.3 (a) SCR control circuit ; (b) rectifier waveform for current control. 



The Unijunction Transistor 425 

resistance load, the output voltage is expressed by 

fJ = ^ (1 + cos0i) (17 - 1) 

For the full-wave rectifier circuit the result is twice as great, or 

£fc = i-(l +COS0J (17.2) 

Mounting, cooling, and temperature limitations for the SCR follow those 
previously discussed for the power transistor. Devices available employ gate 
currents of 50 mA to control operating currents of 50 A or more. 



^ E 



0.6 
























s,(b) 














0.4 










































t-\\ 


*"v 












0.2 






— id) 




s 




















"*• . X 







20" 40° 60° 80° 100° 120° 140° 160° 180° 

Figure 17.4 Control of average rectifier current: (a) half wave, resistance load; 
(b) full wave, resistance load. 



17.3 The Unijunction Transistor (UJT) 

Another trigger device is made with a bar of n silicon having a p junction of 
aluminum-doped material added below the midpoint as shown in Fig. 17.5(a). 
With only one junction, the device has become known as a unijunction 
transistor (UJT). The circuit symbol appears in Fig. 17.5(b). The emitter arrow 
points in the direction of forward current. 

The ends of the bar act as base contacts B, and B 2 . Base B 2 is maintained 
positive at V BB and the bar appears as a resistance of 4000 to 10,000 Q. The 
current through the bar creates a voltage drop along the bar. At the location 
of the p emitter the bar is positive to B, and ground by some fraction of V BB , 
called t\V m . The lowercase Greek letter t] (eta) is called the intrinsic standoff 
ratio. 



426 



Power Control 







°2 



T E 

vVbb 



1 *T 



~x 




(a) (b) (c) 

Figure 17.5 (a) Unijunction transistor construction; (b) circuit symbol; (c) 
performance curve. 

If the voltage v EB applied to the emitter is less than t\V BB , then the emitter 
is negative to the n bar and the junction is reverse-biased. If the voltage v EB 
is made more positive than i\V BB , the emitter becomes forward-biased and 
holes move from the p emitter into the n bar and toward fi, . The presence of 
holes in the bar calls for electrons to enter the region from B x and the 
increased density of mobile charges lowers the resistivity of the bar between 
£and B t . The voltage between £and fl, then drops, allowing more holes to 
enter the bar and results in more electrons entering, rapidly reducing the volt- 
age v EB from V p to that of a normal forward-biased diode, as illustrated in 
Fig. 17.5(c). 

This action can also be explained by use of Fig. 17.5(c) in which the initial 
current at A is only the reverse value I co . The voltage can rise to the peak 
voltage V, where v EB = tjV BB = V p , and emitter forward conduction begins. 
With the entrance of holes into the bar, the bar resistance falls abruptly and 
the voltage drops to C on the forward-bias diode curve. The valley voltage 
V. is the lowest voltage value between emitter and base B,. We thus have a 
device that will rapidly switch from V p to the diode voltage approximating 

r.. 

The value of the peak voltage V „ can be predicted for a given V BB by use 
of 

V, S nV„ + 0.7 V (17.3) 

where the term 0.7 V is an approximation to the inherent forward drop of a 
silicon junction. 

The device yields an equivalent circuit of the form shown in Fig. 17.6. 
The diode is that of the junction; the resistances R, and R 2 are those of the 



' 



Triggering the SCR 



427 



+ V m 






»EB 
O 




Figure 17.6 Equivalent circuit for the UJT. 

two parts of the bar. The value of tj can be found from the resistances of the 
bar as 



1 



(17.4) 



where /?,„ is the value of /?, at l B = 0. The resistance R l0 = 5000 Q and may 
drop to 50 Q when switched by i E , for a typical UJT. 



17.4 Triggering the SCR 

The circuit in Fig. 17.7(a) will provide a variable-phase voltage for trig- 
gering an SCR at a desired angle 6, in the ac cycle. Either R or C may be 
varied to shift the phase of the control voltage V c , derived between points 
B and £ of the phase shift bridge. Voltage drops IR and j/X c = jlj(2nfC) 
must add to V A0 \ this addition must be carried out at right angles because of 
the j coefficient of the reactance term. Then, by geometry, point £ must 
always lie on the semicircular locus with a constant diameter equal to V AD , 
as shown in Fig. 17.7(b). As either R or C is varied, the point £ moves 




"-v E 




(a) 



(b) 



Figure 17.7 (a) Phase shirt control of an SCR; (b) phase shift analysis by circle 
diagram. 



428 



Power Control 



around the circle and voltage V c lags the SCR anode voltage by angle 6, ; 
the magnitude of V c remains constant at one-half of V AD . 

The lag angle 8, may be found from the circuit constants and the geome- 
try of the triangle as 

tan*i = 



- InfCR 



(17.5) 



(17.6) 



l/(2;r/C) 
If R is replaced with an inductor L and C with a resistor, then 

tan*L_^ 

If the range of variation of these components is sufficiently great, the 
angle 0, may be varied nearly from 0° to 180°, resulting in a variation of the 
average value of SCR current as indicated in Fig. 17.4(a). 

Diode D is present to prevent gate current on the negative half cycle and 
resistor R, serves to limit the forward gate current to a safe value. 

A common application of the UJT is in triggering an SCR, as shown in 
the typical circuit in Fig. 17.8. Diode D is incorporated to synchronize the 
gate signal with the positive anode of the SCR. With terminal A and the SCR 
anode positive, the diode blocks current and transistor Q is off. Capacitor 
C is then able to charge through control resistor R„ introducing a lime 
delay in each positive half cycle before the capacitor voltage builds up to 
V p and triggers the UJT. This buildup of voltage is shown from i to /, in 
Fig. 17.8(b) and is determined by the time constant CR X . When the UJT 
triggers, its emitter voltage falls and C is discharged through the UJT and 
R 2 . The sharp pulse of discharge current produces a peaked gate voltage for 
the SCR, as shown between /, and i 2 in Fig. 17.8(c), and the SCR turns on. 
The magnitude of the triggering current for the SCR is controlled by resis- 
tor R 2 . 

The circuit provides a triggering delay in each cycle and also provides 
a pulse waveform for precise triggering of the SCR. The time of triggering 
or the angle 6, is controllable by resistor R x . 

On the negative half cycle at A, diode D supplies voltage to Q and this 
transistor conducts and acts as a short circuit across C. The charging of C 
then starts precisely at the beginning of each positive half cycle at A. 

Resistor R x must vary within limits so that the peak point triggering 
current l„ can be obtained from the I 22 V source. Thus R x must be limited as 



V - V 



£>R* 



(17.7) 



to turn on the UJT. At the valley point the emitter current must fall below /, 
to turn off the UJT: that is, 



(17.8) 



Triggering the SCR 




(a) 



»EB 



Vp~ 




'„ ' 



y 



(b) 



LA 



to 



Figure 17.8 (a) Phase control with a unijunction trigger; (b) and (c) triggering 
waveforms. 

The steady UJT current must not produce a voltage equal to the gate voltage 
across R z ; this places a limit on R 2 . 

Example: A typical UJT is rated 

P t = 300 mW y„ = 35 V 

tsbmk) = 2 A R BB = 7000 ft 

i*(m) = 50 mA tj = 0.6 



430 



Power Control 



Typical circuit values may be chosen as V BB = 30 V, V r — 1 V, /„ = 10 /iA, 
and /, = 10 mA. Using Eq. 17.3, 

y, - 1Vbb + 0.7 

= 0.6 x 30 0.7= 18.7 V 
WithEq. 17.7 

V V. 30 - 18.7 11.3 , ., l06a . R 

With Eq. 17.8, 

V- y. 30-1 29 _ 290 oq</? 

^TT 10 X io-» - To^ - 290 °" < R * 

Therefore we have available the range of 1. 1 3 MQ to 2900 Q for the varia- 
tion of R x , in controlling 9, for triggering the SCR. 



17.5 The Controlled Polyphase Rectifier 

Most industrial power is supplied by three-phase circuits. When dc power is 
needed, it is convenient to employ a polyphase rectifier, one form of which 
is shown in Fig. 17.9. This is a bridge form of circuit, in which two rectifier 
elements conduct in series at a given time. Control of the starting lime of 
current conduction in SCR,, SCR 2 , SCR 3 is furnished by a three-phase 
adaptation of the unijunction transistor circuit shown in Fig. 17.8. Each 
diode conducts at the lime its corresponding SCR turns on. 

More complex circuits are used for providing larger amounts of power. 



i ii> 





Figure 17.9 Three-phase controlled bridge rectifier circuit. 



Shunt-Wound DC Motor Control 



431 



17.6 Shunt-Wound DC Motor Control 

The speed-control circuit for a dc shunt-wound motor illustrates the ap- 
plication of some of the principles previously discussed. A simplified circuit 
is shown in Fig. 17.10, which operates by controlling the triggering time of 
an SCR so as to vary the average voltage applied to the motor armature. 

The UJT circuit is similar to that of Fig. 17.8, with diode D supplying a 
positive voltage to the UJT only on the half cycle in which the SCR anode 
is also positive. Current from D charges C through R x and when the capacitor 
voltage reaches V„ of the UJT, the latter triggers and discharges C through 
R . This sharp pulse of current produces a pulse of gate voltage for the SCR 
and it is triggered on. The current of the SCR is the armature current of the 
motor, occurring in partial half-wave pulses as shown in Fig. 17.10(b). 
Adjustment of R, controls the triggering angle 9, and thereby the average 
current to the motor. The result is variation of motor speed. 



Field 




AC 




(a) 



0, 



(b) 
Figure 17.10 DC motor control circuit; (b) waveform of current. 



432 



Power Control 



^ 



Review Questions 



433 



If ihc motor is heavily loaded, it tends to slow down, the voltage across 
the armature will be reduced, and a larger voltage will appear at diode D 
and across the control circuit. This larger voltage decreases the time required 
to charge C and trigger the SCR and yields an earlier 0, and a longer current 
pulse to the armature. This longer pulse tends to speed up the motor and 
compensates for motor loading. Speed regulation of about 10 per cent is 
possible. 

The motor field current is supplied by a separate bridge rectifier. 



77.7 Comments 

We now have a family of power control switches. Since all units of the family 
are basically similar to the SCR, we have discussed applications only in terms 
of that unit. A listing of the devices seems worthwhile, however, and is pre- 
sented here: 

SCR: Behaves as a rectifier diode, blocking current in the forward 

direction until a current pulse of sufficient magnitude is applied 
to the gate electrode. 

Thyristor: A general name for the entire family of power control switches 
but most usually applied to the device we here call the SCR. 

Triac: A two-way conduction version of the SCR, with current trig- 

gering by a gate pulse. The action of the gate controls the time 
of triggering in both current directions. 

Diac: A two-electrode, three-layer device acting as two inverse-parallel 

diodes. It conducts in either direction after the applied voltage 
exceeds a value called the breakover voltage. 



<$r cr 



(b) 



(a) 



© 



(c) (d) 

Fifcure I7.lt (a) Triac; (b) SCR ; (c) diac; (d) GTO. 







GTO: A gale-turn-off switch; a one-way device with turn-on cha- 

racteristics as in the SCR. By driving a sufficient negative current 
into the gate, the GTO can also be turned off. 

Circuit symbols for these devices are shown in Fig. 17.11. For further 
information, reference should be made to the manuals of several manu- 
facturers, as 

SCR Manual, G.E. Semiconductor Products Dept., Electronics Park, 
Syracuse, N.Y. 

RCA Solid-Stale Power Circuits, SP-52, RCA, Somervillc, N.J. 08876. 
Westinghouse SCR Designer's Handbook, Weslinghouse Corp., Semi- 
conductor Division, Youngwood, Pa. 



REVIEW QUESTIONS 

17.1 What is an SCR? 

17.2 Explain the process of cumulative current buildup when an SCR is gated to 
turn on. 

17.3 What is the typical turn-on time of an SCR? 

17.4 What turn-off lime is expected for an SCR? 

17.5 Explain the SCR action in the different current regions of Fig. 17.2. 

17.6 What is meant by the holding current? 

17.7 How do you turn off an SCR? 

17.8 What is a UJT? 

17.9 What is the intrinsic standoff ratio? 

17.10 How is t] (eta) measured? 

17.11 Explain the process lhat occurs in a UJT when it triggers. 

17.12 What is the peak voltage in a UJT? 

17.13 What is the valley voltage? 

17.14 If C is fixed and R is variable, what are the limits on R\fO t is to vary between 
0° and 180° in a phase shift bridge? 

17.15 What is the function of R, in the gate circuit of an SCR? 

17.16 What factors determine the lime between voltage application and triggering 
of the UJT in the circuit of Fig. 17.8? 

17.17 What is the purpose of transistor Q in the circuit of Fig. 17.8? 

17.18 Why do we not have to use two sets of SCR elements in the polyphase bridge 
rectifier of Fig. 17.9? 

17.19 Explain the method by which speed is controlled in the motor circuit of Fig. 
17.10. 



434 



Power Control 



PROBLEMS 

17.1 The ac supply to an SCR half-wave rectifier is 240 V rms. When the load resis- 
tance is 1000 fi, what are the dc load currents when the SCR is triggered at 
0°, 40°, 90°, and 135°? 

17.2 A 10-J2 load is connected to a 120-V rms supply line through an SCR. The 
average load current is to be varied between 3.0 and 0.5 A. What range of 
triggering angles is needed ? 

17.3 A 10-12 load is supplied with an average current from a 120-0-120 V center- 
tapped transformer through a full-wave SCR rectifier circuit. The load power 
is to be varied between 100 and 40 W. 

(a) What arc the load voltages ? 

(b) What is the range for the triggering angle ? 

17.4 A phase shift bridge at 60-Hz is to control an SCR from full on to 20 per 
cent of full-on current. If C = 0.7 //F, find the range of variation of R. 

17.5 A phase shift bridge at 60-Hz is to control an SCR from 10 per cent to full-on 
current. The value of C = 0.15 fiF. What is the range needed for /?? 

17.6 In Fig. 17.12, R, = 5000 Jl, C, = 0.5 /*F, and R L - 60 fi. What is the 
angle 0,, and what is the average load current /„ for V ■ 60 V? 

17.7 In Fig. 17.12, /?, = 2500 O. and R L - 10 fi. What value of C, will cause 
the SCR to trigger at 50°? 




SCR 



Figure 17.12 



17.8 An SCR is used to control the power into a 1000-W, 57-fi, heater element 
from a 540-V rms circuit. What triggering angles should be used for |, J, and 
I rated heat ? 

17.9 A UJT has R l0 - 5000 Q and /? 2 = 6000 Q. What is the value of rjl With 
V BB = 25 V, what is the expected peak voltage? 



Index 



^4/3 plot, tabic. 219 
Acceptor atom, 10 
Active devices, 48 
Active network, 50 
Adder circuit, 399 
Addition, binary, 382 
a (alpha), 70 
Amplification factor: 
transistor, 70 
triode, 148 
Amplifier: 
band limits, 167 
bandpass, 261 
bass boost in, 191 
buffer. 322 
cascodc, 252 
differential, 228 

common-mode rejection, 232 
direct-coupled, 225 
FET: 
basic circuits, 132 
common-drain, 137 
common-source, 136 
design of, 140 



Amplifier (com.) 
FET (com.) 

low-frequency response, 175 

source resistor, 179 
gain of a cascade, 59 
gain stability of, with feedback, 217 
normalized gain of, 168 
operational, 238 

as comparator, 245 

definitions, 249 

for differentiation, 244 

frequency compensation, 246 

for integration, 243 

inverting, 238 

millivoltmetcr, 245 

noninverling, 240 

offset voltage and current, 249 

RC amplifier, 323 

scaling in, 244 

single-ended output, 250 

slew rate, 247 

summing, 242 

unity-gain isolator, 242 
pentode, 154 
phase angle of, 168 
power, 283 



435 



436 



Index 



Index 



437 



^ 



Amplifier (com.) 
power (com.) 
Class A, 286 
Class AB, 298 
Class B, 298 
Class B linear. 307 
Class C, 284, 333 
conversion efficiency of, 288 
distortion of, 293 
output transformer-less, 302 
thermal environment, 290 
push-pull, 297 
Class B, 300 
stagger-tuned, 271 
transistor: 
A, to A v relation, 81 
basic forms, 74 
bias-stabilized, 107 
cascaded, 164 
C-E, 82 

at high frequency, 188 
C-C. 85 
C-E, 78 

equivalent circuit of, 75 

design of, 107 
comparison of performance, tabic, 
complementary symmetry, 306 
differential, 228 

constant-current bias for, 236 

single-ended, 231 
emitter follower, 85 

design, 1 14 
fixed bias, 100 

design, 105 
four-resistor bias, 102 

design, 107 
frequency response, 164 
high-frequency response, 182 
integrated, 225 
low-frequency response, 168 
pass band, 165 
performance, table, 87 
phase inverter, 304 
single-tuned, 268 
stability ratio, 105; table, 115 
treble boost in, 192 
voltage-feedback bias for, 1 10 
voltage reference for, 237 
triode: 
basic circuits, 148 
cathode follower, 155 



Amplifier (conl.) 
triode (conl.) 
cathode resistor, 179 
common-cathode, 153 
grounded-grid, 157 
low-frequency response, 175 
video, 261 
scrics-pcakcd, 277 
shunt-peaked, 273 
time delay, 275 
Anode, 21 

vacuum tube, 145 
Armstrong, E.H., 367 
Armstrong oscillator, 318 
ASCII code, 379 
Astable multivibrator, 400 
Atoms, 2 
acceptor, 10 
Bohr theory, 2 
donor, 9 
Automatic frequency control, 348 
Automatic volume control, 336 

B 

Balanced modulator, 338 
Bandwidth: 

AM signal, 330 

cascaded amplifier, 189 

definition, 249 

FM signal, 342 

inprovement by feedback, 204 

information, 355 

resonant circuit, 265 
Bardeen and Brattain, 64 
Base, 65 
Bass boost, 191 
BCD code, 378 
Beat frequency, 358 
Bel, 58 

Bell, Alexander Graham, 58 
Bias: 

constant-current, 256 

FET, 127 

fixed, 100 

four-resistor circuit, 129 

stability, 105 

stabilized, design for, 107 

transistor, 96 

voltage feedback, 1 10 
circuit, 112 



Bias currents, 73 

Binary addition, 382 

Binary arithmetic, 377 

Binary code, 374, 378 

Binary-coded-decimal, 378 

Binary number, 374 

Binary point, 376 

Binary signals, decoding of, 413 

Bistable multivibrator, 400 

Bit, 354 

definition, 376 

parallel and serial processing of, 381 
Black box, 48 

feedback concepts, 199 
Bohr atom theory, 2 
Breakover voltage, diac. 432 
Bridge rectifier circuit, 25 
BufTer amplifier, 322 
BVooo, 122 
Byte, 379 



Capacitor: 

blocking, 166.317 

bypass, 151, 177 

filter, 27 

speed-up, 392 
in flip-flop, 402 

voltage-variable, 17 
Capture effect, 367 
Carrier frequency. 327 
Carry bit, 400 
Cascaded amplifier, 189 
Cascodc amplifier, 252 
Cathode, 21 

vacuum tube, 145 
Cathode follower, 148. 155 
Cathode-ray oscilloscope, 370 
Cathode-ray tube, 159 
Center frequency, 342 
Channel, 120 
Clamper, 41 
Class A operation, 284 
Class AB operation, 298 
Class B amplifier: 

linear, 307 

operation, 284 
Class C amplifier: 

modulated, 333 

operation, 284 



Clipper, 38 

double diode, 41 
Clock, 381 

trigger, in RS flip-flop, 403 
CMOS: 
JK flip-flop, 406 

logic circuits, 395 
manufacture, 398 
CMRR, 233, 250 
Code, cxcess-3, 421 
Coefficient of coupling, 269 
Cold-cathode emission, 147 
Collector, 65 
Collector resistance, 68 
Colpitis oscillator, 315 
Common-cathode amplifier: 

circuit, 148 

pentode, 154 

triode, 153 
Common-drain amplifier, 137 

design, 140 
Common mode, definition, 232 
Common-mode input, 228 
Common-mode rejection ratio, 233, 250 
Common-mode voltage gain, 249 
Common-source amplifier, 1 36 
Comparator, 245 
Complement, 377 

Complementary symmetry circuit, 306 
Conductance, 77 
Conduction, 5 
Constant-current bias, 236 
Control generator, 56 
Conversion gain, 340 
Covalcnt bond, 5, 8 
CRf, 29 

Critical coupling, 269 
Crossover distortion, 306 
Crystals. 3. 10 

piezoelectric, 319 
Current: 

definition, 1,7 

diode, 13, 14 

forward, 12 

h- bo, 67 

in injunction, 13 

reverse saturation, 1 3 
transistor, 67 

transistor, table, 99 
Current amplification factor, 70 

relation with A„ 81 











Index 




439 


438 




Index 




Equivalent circuit (com.) 


Filter (com.) 




Current division factor, 170 


Diode-transistor logic (DTL), 390 






Thevenin, 51 


capacitor (com.) 




Current feedback, 207 


Direct-coupled amplifier, 225 




transistor: 


surge resistor for, 31 




Current generator, controlled, 56 


Directly-heated filament, 147 






g m model, 78 


C/?/ factor in. 29 




Current-source equivalent circuit, 52 


Discriminator, frequency, 344 
correction voltage from, 368 






high-frequency, 181 
/j-parameter, 75 


n (pi), 31 
n-R, 34 




D 


Dissipation locus, 97 






hybrid-K, 180 


SSB signal. 362 





Darlington compound transistor, 250 

Dc level shifter, 237 

Decade, frequency, 168 

Decibel, 58 

Decimal counting, 414 

Decimal numbers, 375 

Decimal point, 375 

Decoding matrix, 413 

Deletion region, 13 

in FET, 121 
Detection, 326 

AM: 
linear, 334 
product, 340 

FM: 
discriminator, 344 
ratio detector for, 347 

radar, 370 
Deviation ratio, 342 
Diac, 432 
Differential amplifier, 228 

as comparator, 245 
Differential-mode voltage gain, 250 
Differentiation, 244 
Diffusion, 11, 89 
Digital circuits, 374 
Diode: 

breakdown of, 15 

capacitance of, 1 7 

clamper, 41 

clipper, 38 
double, 41 

current in, 14, 23 
equation, 13 

ideal, 21 

junction, 1,11 

light-emitting, 16 

photo, 16 

rectifier, 13, 21 

steering, 403 

symbol for, 14 

Zener, 15 
Diode modulator, 331 

balanced, 339 



Distortion: 

allowable, 296 

crossover, 306 

frequency, 165 

nonlinear, 206, 293 

reduction by feedback, 206 
Divide-by- 10 counter, 416 
Donor atom, 9 
Dopplcr effect, 370 
Drain, 120 
DSB, 330 



Edison effect, I 

EHF, 371 

Electron, 1,2 

Electron gun, 159 

Electron-hole pair, 7 

Electron volt, 8 

Element, 2 

Emission of electrons, 146 

Emitter, 65 

Emitter-coupled logic (ECL), 394 

Emitter follower, 85 

circuit design, 1 14 

feedback amplifier, 215 

phase inverter, 306 
Emitter resistor, 177 
Energy: 

forbidden, 7 

gap, 7 
Enhancement mode, 1 25 
Epitaxial growth, 91 
Equivalent circuit: 

C-B amplifier, 83 

C-C amplifier, 85 

C-E amplifier, 75 

current source, 52 

definition, 49 

FET, 135 
high-frequency, 181 

//-parameter, 56 

Norton, 52 

SCR, 422 



triode, 152, 182 
UJT, 426 
voltage-source, 51 
Exccss-3 code, 421 



Fan-in, 388 
Fan-out, 388 
Feedback: 

advantages, 221 

bandwidth improvement. 204 

black-box analysis, 199 

closed-loop gain, 201 

control of resistances. 207 
table, 209 

current, 207 

current-series, 209 

in decimal counter, 415 

emitter follower, 215 

gain margin in, 220 

gain stability with, 217 

internal gain with, 200 

measure of, 201 

multiple-stage, 216 

oscillator, 201 

phase margin in, 220 

positive, 201 

limitation in receiver, 359 

principles of, for oscillators, 313 

reduction of distortion. 206 

stabilization of gain, 202 

voltage, 207 

voltage-series, FET, 213 

voltage-shunt, 212 
Feedback factor, 201 
FET, 120 
Figure of merit : 

Q, 263 

transistor, 186 
in C-B circuit, 189 
Filament, 147 
Filter: 

capacitor, 27 



First breakdown, 290 
Flip-flop. 400 

decimal counting, 414 

shift register, 412 
Fluorescent screen, 161 
FM system, 367 

pre-emphasis circuits in, 193 
Forbidden energy, 7 
Foster-Seeley discriminator, 345 
Frequency classification, table, 371 
Frequency compensation, 246 
Frequency deviation, 341 
Frequency distortion, 165 
Frequency emphasis, 191 

insertion loss with, table. 192 
Frequency modulation, 341 
Frequency multiplex, 362 
Frequency translation, 360 
Full-wave rectifier circuit, 24 



Gain-bandwidth product, 187 
Gain, voltage, 77 
Gain margin, 220 
Gap energy, 8 
Gate: 
FET, 120 
logic, 383 
and. 383 
diode, 386 
fan-in of, 388 
fan-out of. 388 
nand, 385 
nor, 385 
not, 384 
or, 384 
Gate-turn-off switch, 433 
Germanium, 2, 5, 8 
Glitch, 394 

Km- 

FET, 133 
transistor, 77 
triode, 150 



440 



Index 



Index 



441 



Grid, 145, 151 
Grounded-grid amplifier, 157 

H 

Half-power frequency, 167 

Half-wave rectifier circuit, 22 

Hartley oscillator, 315,316 

Hartley-Shannon law, 357 

Heat sink, 291 

Heterodyne interference, 358 

Hexadecimal system, 381 

HF, 371 

Holding current, 424 

Hole, 6 

h parameters: 

conversion, 81 

current gain, 70 

definitions, 55 

equivalent circuit, 56 
Hybrid parameters, 56 
Hybrid-n equivalent circuit, 180 
Hydrogen atom, 2 



I 



Icbo, 67 

as temperature function, 99 
Ideal transformer, 285 
Image frequency, 360 
Impurities, semiconductor, 8 
Indirectly-healed cathode, 197 
Information, 354,355 

Hartley-Shannon law for, 357 
Injection of charges, 12 
Input offset current, 250 
Input offset voltage, 250 
Insertion loss, table, 192 
Insulator, 4 
Integrated circuit: 

CMOS, 395 

logic, 389 

manufacture of, 253 

passive elements in, 255 

sheet resistance of, 255 

silicon, 253 
Integration, 243 

Intermediate frequency, 262, 358 
Intrinsic conduction, 7 
Intrinsic standoff ratio, 425 
Inverter, 383 

CMOS, 395 



Ion, 3 

Ionizing energy, 3 



JFET, 120 
JK flip-flop. 405 
Junction, 1 1 
voltage-current equation for, 13 



Lattice, 3 
LED, 16 
LF, 371 
Limitcr, 344 

emitter-coupled, 346 
Load line: 

ac, 288 

dc. 72 
forFET. 125 
Logic circuits, 383 

adder, 399 

CMOS, 395 
inverter, 395 
manufacture, 398 
nand, nor, 396 

diode (DL). 386 

diode-transistor (DTL), 390 

emitter-coupled (ECL), 394 

integrated, 389 

resistance-transistor (RTL), 392 

transistor-transistor (TTL), 393 

voltage levels in, 386 

M 

Majority carriers, 10 

Masking, 90 

Matched load, 54 

Materials, properties of, table, 4 

Matrices, decoding, 413 

Maximum power output, 53 

Metal, 4 

Metric magnitudes, 4 

MF, 371 

Miller effect, 180 

Millivoltmctcr circuit, 245 

Minority carriers, 10 

Modulated amplifier, 333 



Modulation, 326 
amplitude (AM). 327 
bandwidth for, 330 
beat frequency. 358 
carrier frequency, 327 
frequency spectrum, 327 
heterodyne interference in, 

367 
"monkey chatter" in, 367 
power spectrum of, 330 
side frequencies, 328 
single sideband, 337 
comparison of AM and FM, 349, 

366 
frequency (FM), 327 
bandwidth, 342 
capture effect, 367 
center frequency, 342 
deviation ratio, 342 
frequency deviation in, 341 
frequency spectrum, 341 
generation, 344 
interference in, 366 
narrow band, 344 
wide band, 344 
phase (PM), 327 
pulse, 327 
Modulation factor, 328 
Modulator: 
balanced, 338 
Class B, 334 
diode. 331 
Monkey chatter, 367 
Monostablc multivibrator, 400 
MOSFET, 123 
Motor control, 431 
Multiplex, frequency, 362 
Multivibrator, 374,400 
astablc, 408 
bistable, 401 
flip-flop, 401 
CMOS, 406 
JK, 405 
RS circuit, 402 
scalc-of-four, 404 
sale-of-two, 403 
toggling (T), 403 
free-running, 408 

synchronization of, 410 
monostablc, 407 
Schmitt trigger, 411 



N 

nand circuit, 391,396 

Negative feedback, 199 (see Feedback) 

Negative logic, 386 

Network : 

active, 50 

current-source equivalent, 52 

equivalent, definition, 49 

maximum power transfer, 53 

Norton, 52 

one-port, 49 

^-matching, 308 

Thevenin, 51 

two-port, 49 
/; parameters for, 54 

voltage-source equivalent, 51 
Neutron, 2 
Noise, 353 

in logic circuits, 394 
Noise figure, 354 
nor circuit, 396 
Normalized gain, 168 
Norton circuit, 52 
not operation, 383, 388 
Number systems, table, 380 
Nyquist criterion, 219 



Octal system, 379 
Offset current, 249 
Offset voltage, 249 
One-shot multivibrator. 407 
Operational amplifier, 225, 238 

comparator, 245 

definitions, 249 

differentiation with, 244 

frequency compensation, 246 

integration with, 243 

inverting, 238 

millivoltmeter, 245 

noninverting, 240 

offsets, 249 

slew rate, 247 

summing, 242 

unity-gain isolator, 242 
Oscillator: 

Armstrong, 318 

astablc multivibrator form of, 408 

Colpitts, 315 



442 



Index 



Index 



443 



Oscillator (com.) 

drain-tuned FET, 322 

Hartley, 315,316 

piezoelectric control of, 319 

principles of, 313 

resistance-capacitance feedback, 322 
Output circuit, without transformer, 302 
Overshoot, 271 



Pair generation, 7 

PAM, pulse-amplitude modulation, 327 
Parallel processing of data, 381 
Parallel-resonant circuit, 263 

bandwidth, 265 

as a tank circuit, 317 
Parameters: 

//. 54 

conversion of, 74 

small-signal, 74 
Passive elements, 255 
PCM, pulse-code modulation, 327 
PDM, pulse-duration modulation, 327 
Peak reverse voltage, 23 
Pentode, 145. 150 
Phase inverter, 304 

complementary symmetry form, 306 

emitter-coupled, 231 
Phase margin, 221 
Photodiode, 16 
Photoelectric emission, 147 
Photoisolator, 16 
Photon, 1 

Photoresist lacquer, 90 
n (pi) filter, 31 
n-matching network, 308 
n-R filter, 34 
Piezoelectric crystals, 319 
Pinch-off, 121 
PMMC instrument, 37 
Polyphase rectifier, 430 
Port, 48 

Position, in number system, 375 
Positive logic, 386 
Power : 

control of. 422 

decibel, 58 

maximum, 53 
Power gain, definition, 79 
Power ratio, in dB, 59 



Prc-cmphasis circuit, 193 
Product detector, 340 
Proton, 2 
PRV, 23 
Pulse, 271 
Pulse stretcher, 407 
Punch-through, 290 
Push-pull circuit, 284. 297 
phase inverter for, 304 



Q, circuit, 263 
Q point, sec Quiescent point 
Quartz crystal, 319 
Quiescent point, 73 

choice of, 96 

Class fl, 300 

variation of, 99 



R 



Radar, 368 

Radio-frequency amplifier, 261 

Radio-frequency choke, 3 1 7 

Radio signal, classification of frequencies, 

371 
Radio systems, 353 
Radix, 374 
Ratio detector, 347 
RC amplifier, 164 
Real time, 327 
Receiver: 

FM, 368 

radar, 370 

SSB, 363 

superheterodyne, 358 
Rectifier, 21 

diode, 13 

silicon-control (SCR), 422 
triggering of, 427 
Rectifier circuit: 

bridge, 25 

full-wave, 24 

half-wave, 22 

polyphase, 430 

SCR, 424 

voltage-doubling, 35 
Rectifying ac voltmeter, 37 
Repetition rate, 272 

in radar, 370 







Reset-set (RS) flip-flop, 402 

Resistance-transistor logic (RTL), 392 

Resistivity, 4 

Resonance, 263 

Reverse saturation current, 13 

transistor, 67 
Ripple factor, 27 
Rise time, 271 
Rochclle salt, 320 
RS flip-flop, 402 



Saturation line, 68 
Scale factor, 244 
Scalc-of-four circuit, 404 
Scale-of-two circuit, 403 
Schmitt trigger, 41 1 
Secondary emission, 147 
Second breakdown, 290 
Selectivity, 261 
Semiconductor: 

atomic arrangement in, 2 

defined, 5 

impurity, 8 

majority carriers, 10 

minority carriers, 10 

resistivity of, table, 4 
Series processing of data, 381 
Series resonance, 320 
Shape factor, 262 
Sheet resistance, 255 
Shell, energy, 2 
Shift register, 412 
Shockley, 120 

Short circuit limit frequency, 1 86 
Sidebands, 329 
Side frequencies, 328 
Signal-to-noise ratio, 354 
Silicon, 5 

atomic arrangement, 2 

forbidden energy, 8 

integrated circuit, 227 
Silicon-control rectifier (SCR), 422 

in motor control, 431 

polyphase circuit, 430 

triggering of, 427 
Single sideband system, 337 

advantages, 366 

receiver, 360 

transmitter, 362 



Skirt selectivity, 360 
Slew rate. 247 

definition, 250 
Small signal, defined, 98 
Source, 50 

FET, 120 
Source follower, 132,137 
Speedup capacitance, 392 
Square wave, harmonics in, 

table, 273 
SSB, 337 
Stabilizing ratio, 105 

for C-E amplifier, 110 

comparison, table, 115 
Stagger-tuned amplifier, 271 
Static, 353 
Steering diodes. 403 
Summing operation. 242 
Superheterodyne receiver, 358 

Tor FM, 368 

for radar, 371 
Surge resistor, 31 
Sweep voltage, 1 59 
Switch, 18,21 

Switching time, flip-flop, 402 
Synchronization, of multivibrator, 410 



Tank circuit, 317 

Tape recording, 193 

Temperature, absolute zero, 6 

T flip-flop, 403 

Thermal derating curve, 292 

Thermal noise, 353 

Thermal resistance, 291 

Thermionic emission, 146 

Thevenin equivalent circuit, 51 

Thyristor, 432 

Time delay, 275 

Total harmonic distortion, 296 

Totem pole, 393 

Transceiver, 363 

Transconductancc : 

conversion, 340 

FET, 133 

transistor, 77 
C-B connection, 83 

triodc, 150 
Transfer curves, 228 

FET, 133 



Index 



Transformer: 
double-tuned, 269 
ideal, 285 
Transistor, 64 
alloyed junction, 89 
bipolar, 64, 120 
CMOS, 395 
complementary, 306 
current relations, table, 99 
Darlington compound, 250 
defined currents and voltages, 64 
derating curve for, 292 
epitaxial growth, 91 
field effect (FET), 64,120 

bias, 127 

characteristics, 132 

dcload line, 125 

depletion mode, 121 

enhancement mode, 125 

equivalent circuit for, 135 

high-frequency model, 181 

junction, 120 

MOS, 123 

pinch-off region in, 121 

symbols, 125 

iransconductance, 133 

transfer curve, 133 

variable resistance, 142 
figure of merit, 186 
frequency limits, 185 
gain-bandwidth product, 187 
grown junction, 89 
high-frequency model, 181 
hybrid-* model, 180 
junction, 65 

manufacturing techniques, 89 
mesa, 89 

operating region, 96 
planar, 89 

reverse saturation current, 67 
saturation line, 68 
thermal environment, 290 
transconductance, 77 
unijunction (UJT), 425 

for motor control, 431 
unipolar, 64, 120 
volt-ampere curves, 68 
voltage limitations, 290 
Transistor amplifier: 
basic types, 74 
bias circuit design, 130 



Transistor amplifier (com.) 

fixed bias for, 100 

four resistor circuit, 102 
FET, 129 

in not operation, 388, 390 

voltage feedback with, 123 
Transistor-transistor logic (TTL), 393 
Translation, frequency, 326, 329, 339 
Transmitter: 

AM, 357 

FM, 367 

radar, 370 

SSB, 362 
Treble-boost circuit, 192 
Triac, 432 
Trigger circuit: 

for SCR, 427 

Schmitt, 411 

trailing edge, 405 
Triodc, 145 

amplification factor, 148 

characteristics, 148 

equivalent circuit, 152 

gain-bandwidth product, 187 

high-frequency model, 182 

notation, 145 

planar, 146 

symbol, 147 

transconductance, 150 
Turnover frequencies, 191 

in FM, 193 
Turns ratio, 285 

U 

UHF, 371 

Undcrcoupling, 269 
Unijunction transistor (UJT), 425 

in motor control, 431 

for triggering, 428 
Unity gain isolator, 242 



Vacuum tube, 145 
Valence electron, 2 
VHF. 371 

Video amplifier, 261 
Voltage, UJT: 

peak, 426 

valley, 426 



Index 



445 



Voltage-doubling rectifier circuit, 35 
Voltage feedback, 207 

bias, 110,112 
Voltage generator, controlled, 56 
Voltage multiplier, 37 
Voltage reference, 237 
Voltage regulator, Zencr, 15 
Voltage-source equivalent, 51 
Voltage-variable capacitor (VVC), 17 



Voltage-variable resistance (VVR), 142 
Voltmeter: 

peak, 38 

rectifying, 37 



Zencr diode, 1 5 
Zero level, 59